Jekyll2023-05-12T15:25:57+00:00https://nickpgill.github.io/feed.xmlNick GillRefugees are welcome!2021-09-30T00:00:00+00:002021-09-30T00:00:00+00:00https://nickpgill.github.io/refugees-are-welcome<p>This post is about an interview I didn’t listen to and a book I didn’t read. Well, sort of. Really it’s about me wanting to say loud and clear that I want my country to do the right thing and welcome asylum seekers into our country to make their new home.</p>
<p>So, to the interview: it was the TODAY programme last week some time. The host was welcoming Alexander Downer to the programme to talk about Australia’s policy on migrants and refugees. I know that name, Alexander Downer, well: he was a minister in the Howard government when I was growing up in Australia in the late 90’s. It’s not a happy memory – I turned the radio off.</p>
<p>I didn’t need to hear Alexander Downer explaining his involvement in the immigration debate because I remember it well. The Howard government, Downer included, used immigration as an election tactic: they whipped up fear about people arriving on boats and they demonstrated their tough-guy credentials by refusing to let people land. This led, ultimately, to subsequent Australian Governments setting up detention centres (read: concentration camps) on nearby islands, most notably, most infamously, on Manus Island and on Nauru.</p>
<p>Politically this policy was brilliant. The Howard government romped to election victory after election victory. Howard (and Downer) were at the helm of government in Australia for a decade and the country became a different place because of it – being “tough on migrants” has become the norm across the Australian political spectrum. No wonder this policy is being looked at admiringly by our current Tory masters.</p>
<p>Ethically, though, this policy was appalling. In human terms it was devastating. People who had hoped to seek asylum in the richest country in the region instead spent years in appalling conditions in detention. Self-harm and suicide occurred in these centres at <a href="https://www.theguardian.com/australia-news/2017/may/18/self-harm-suicide-and-assaults-brutality-on-manus-revealed">incredible levels</a>, sexual abuse <a href="https://www.theguardian.com/australia-news/2016/aug/10/the-nauru-files-2000-leaked-reports-reveal-scale-of-abuse-of-children-in-australian-offshore-detention">likewise</a>. Perhaps most appalling of all, many of the people suffering these brutalities were <a href="https://www.bbc.co.uk/news/world-asia-45327058">children</a>. The policy continues despite <a href="https://www.hrw.org/news/2021/07/16/australia-8-years-abusive-offshore-asylum-processing">fierce criticism</a>.</p>
<p>THIS is the future that Priti Patel evisions for the UK. It is a future where the UK ducks its obligations under international law; it is a future where innocent people who need our help are, instead, subjected to detention and abuse. It is a future that, in some ways, <a href="https://www.politics.co.uk/reference/immigration-removal-detention-centres/">has already arrived</a>. It is not the future that I want for my country.</p>
<p>I am writing this because it is easy to listen to radio interviews with (apparently) reasonable people and to be taken in by their sober tone, their claims of “the reality of the world today”, their talk of “difficult decisions”. It seems important to me to state very clearly that “I disagree”. In fact these decisions are not so difficult at all – we are a wealthy country and we have more than enough resources to help the people who need it. The only decision we need to make is to commit to honouring our obligations under international law and to providing the help and assistance that people need to get themselves on their feet. After this, we can feel certain that the contribution they make to our country will be a benefit to us all – not that this is the reason for welcoming them in their hour of need, it is just an added bonus.</p>
<p>Let me emphasise one point: I mentioned international law just now and it pertains directly here because Australia has been criticised by the UNHCR on a number of occasions for violating their legal obligations in international law. Australia’s response has been to <a href="https://www.unhcr.org/en-au/5ef99e0a4.pdf">repeatedly double-down</a>. It is despicable that such a law-breaking regime should be considered a valid model for our own government to emulate – despicable, but hardly surprising given the people in charge in the UK right now. It is perhaps more of an aberration, though, that the BBC considers this a legitimate question to debate, to the point that they would invite an Australian politician to extol the virtues of law-breaking on the BBC’s flagship political programme. I would argue that the only platform Downer should be given to discuss Australian immigration policy is in the dock at The Hague.</p>
<p>And so to the book I haven’t read: It is called “No friend but the mountains” and was written by Behrouz Boochani. Boochani was illegally detained on Manus Island for a number of years; the book was written during this time on his mobile phone, using his thumb, text by text, and sent out via Whatsapp. The book describes this ordeal. It is on my desk as I write; I expect it to be a difficult read.</p>
<p>One line from the foreword has struck me: the writer Richard Flanagan discusses the politics of immigration in Australia and speaks of “policies in which both our major parties have publicly competed in cruelty”. This line sums it up beautifully (and awfully ) – this has indeed been the story in Australia for the last 25 years. We need to make damn sure that it isn’t the story in the UK for the next 25.</p>nickgillThis post is about an interview I didn’t listen to and a book I didn’t read. Well, sort of. Really it’s about me wanting to say loud and clear that I want my country to do the right thing and welcome asylum seekers into our country to make their new home. So, to the interview: it was the TODAY programme last week some time. The host was welcoming Alexander Downer to the programme to talk about Australia’s policy on migrants and refugees. I know that name, Alexander Downer, well: he was a minister in the Howard government when I was growing up in Australia in the late 90’s. It’s not a happy memory – I turned the radio off. I didn’t need to hear Alexander Downer explaining his involvement in the immigration debate because I remember it well. The Howard government, Downer included, used immigration as an election tactic: they whipped up fear about people arriving on boats and they demonstrated their tough-guy credentials by refusing to let people land. This led, ultimately, to subsequent Australian Governments setting up detention centres (read: concentration camps) on nearby islands, most notably, most infamously, on Manus Island and on Nauru. Politically this policy was brilliant. The Howard government romped to election victory after election victory. Howard (and Downer) were at the helm of government in Australia for a decade and the country became a different place because of it – being “tough on migrants” has become the norm across the Australian political spectrum. No wonder this policy is being looked at admiringly by our current Tory masters. Ethically, though, this policy was appalling. In human terms it was devastating. People who had hoped to seek asylum in the richest country in the region instead spent years in appalling conditions in detention. Self-harm and suicide occurred in these centres at incredible levels, sexual abuse likewise. Perhaps most appalling of all, many of the people suffering these brutalities were children. The policy continues despite fierce criticism. THIS is the future that Priti Patel evisions for the UK. It is a future where the UK ducks its obligations under international law; it is a future where innocent people who need our help are, instead, subjected to detention and abuse. It is a future that, in some ways, has already arrived. It is not the future that I want for my country. I am writing this because it is easy to listen to radio interviews with (apparently) reasonable people and to be taken in by their sober tone, their claims of “the reality of the world today”, their talk of “difficult decisions”. It seems important to me to state very clearly that “I disagree”. In fact these decisions are not so difficult at all – we are a wealthy country and we have more than enough resources to help the people who need it. The only decision we need to make is to commit to honouring our obligations under international law and to providing the help and assistance that people need to get themselves on their feet. After this, we can feel certain that the contribution they make to our country will be a benefit to us all – not that this is the reason for welcoming them in their hour of need, it is just an added bonus. Let me emphasise one point: I mentioned international law just now and it pertains directly here because Australia has been criticised by the UNHCR on a number of occasions for violating their legal obligations in international law. Australia’s response has been to repeatedly double-down. It is despicable that such a law-breaking regime should be considered a valid model for our own government to emulate – despicable, but hardly surprising given the people in charge in the UK right now. It is perhaps more of an aberration, though, that the BBC considers this a legitimate question to debate, to the point that they would invite an Australian politician to extol the virtues of law-breaking on the BBC’s flagship political programme. I would argue that the only platform Downer should be given to discuss Australian immigration policy is in the dock at The Hague. And so to the book I haven’t read: It is called “No friend but the mountains” and was written by Behrouz Boochani. Boochani was illegally detained on Manus Island for a number of years; the book was written during this time on his mobile phone, using his thumb, text by text, and sent out via Whatsapp. The book describes this ordeal. It is on my desk as I write; I expect it to be a difficult read. One line from the foreword has struck me: the writer Richard Flanagan discusses the politics of immigration in Australia and speaks of “policies in which both our major parties have publicly competed in cruelty”. This line sums it up beautifully (and awfully ) – this has indeed been the story in Australia for the last 25 years. We need to make damn sure that it isn’t the story in the UK for the next 25.Black Lives Matter2020-06-02T00:00:00+00:002020-06-02T00:00:00+00:00https://nickpgill.github.io/black-lives-matter<p>This is a maths blog ordinarily, but there are extraordinary events happening in the USA just now, and I want to offer my support to the protestors. I don’t do the social media thing so won’t be tweeting out and what-have-you, but I can, at least, show my support here.</p>
<p>The situation in the US is long-standing and terrible, and worthy of protest. Here in the UK there is still institutional racism in many parts of society, as well as casual racism in many areas of life. We must be relentless in our efforts to end racism in all its forms.</p>
<p>On Saturday I’ll be going to a gathering in Caerphilly (covid guidelines allowing), to show my support for the fight against racism. I’m committed to supporting my BAME colleagues at work however I can, as well as combatting racism outside of work, in daily life.</p>
<p>Of course, I must also be aware that I’ve grown up in a society tainted by racism, and I can’t help but have been affected by it. I’m also committed to examining my own thoughts, ideas and behaviour to make sure I have not unwittingly internalised racism (or other forms of bigotry). The struggle continues.</p>
<hr />
<p>Addendum: I’ve read a bunch of books on the subject of racism and thought I’d record some below, in case they are of interest:</p>
<ul>
<li><em>Soledad Brother: The prison letters of George Jackson</em></li>
<li><em>The autobiography of Malcolm X</em> by Malcolm X and Alex Haley</li>
<li><em>Roots</em> by Alex Haley</li>
<li>Any of James Baldwin’s books</li>
<li>Any of Toni Morrison’s books, especially <em>Beloved</em></li>
<li>For am Australian angle, Sally Morgan’s <em>My Place</em></li>
<li>More controversially, perhaps: <em>To Kill a Mocking Bird</em> and the follow-up <em>Go Set a Watchman</em> by Harper Lee; the latter book is not as famous to TKAMB, but I found it very insightful as to the mindset of people in the deep South of America. This mindset isn’t always what I want to hear, but if we are to change society, we have to start from where we are not from where we would like to be.</li>
<li>If you don’t have the time just now to read a whole novel, then the great Gil Scott-Heron was able to get his point across <a href="https://www.youtube.com/watch?v=goh2x_G0ct4">in just over 2 minutes</a>… J.B. Lenoir needed <a href="https://www.youtube.com/watch?v=TtzIy1YUVXU">just over 3</a>…</li>
</ul>nickgillThis is a maths blog ordinarily, but there are extraordinary events happening in the USA just now, and I want to offer my support to the protestors. I don’t do the social media thing so won’t be tweeting out and what-have-you, but I can, at least, show my support here. The situation in the US is long-standing and terrible, and worthy of protest. Here in the UK there is still institutional racism in many parts of society, as well as casual racism in many areas of life. We must be relentless in our efforts to end racism in all its forms. On Saturday I’ll be going to a gathering in Caerphilly (covid guidelines allowing), to show my support for the fight against racism. I’m committed to supporting my BAME colleagues at work however I can, as well as combatting racism outside of work, in daily life. Of course, I must also be aware that I’ve grown up in a society tainted by racism, and I can’t help but have been affected by it. I’m also committed to examining my own thoughts, ideas and behaviour to make sure I have not unwittingly internalised racism (or other forms of bigotry). The struggle continues. Addendum: I’ve read a bunch of books on the subject of racism and thought I’d record some below, in case they are of interest: Soledad Brother: The prison letters of George Jackson The autobiography of Malcolm X by Malcolm X and Alex Haley Roots by Alex Haley Any of James Baldwin’s books Any of Toni Morrison’s books, especially Beloved For am Australian angle, Sally Morgan’s My Place More controversially, perhaps: To Kill a Mocking Bird and the follow-up Go Set a Watchman by Harper Lee; the latter book is not as famous to TKAMB, but I found it very insightful as to the mindset of people in the deep South of America. This mindset isn’t always what I want to hear, but if we are to change society, we have to start from where we are not from where we would like to be. If you don’t have the time just now to read a whole novel, then the great Gil Scott-Heron was able to get his point across in just over 2 minutes… J.B. Lenoir needed just over 3…Jan Saxl2020-05-10T00:00:00+00:002020-05-10T00:00:00+00:00https://nickpgill.github.io/jan-saxl<p>A week ago I heard the very sad news that Jan Saxl has died. Jan was my PhD supervisor at Cambridge – I spent three and a half years working under his guidance. The opportunity to work with Jan was one of the great privileges of my mathematical life, and I owe him a great debt for helping me through the difficult process of researching and writing a doctoral thesis.</p>
<p>As well as being a fine mathematician, Jan was a warm and witty man. Now I have PhD students of my own I know that it is not always easy to know what the right thing to do is when it comes to supervision – looking back I appreciate Jan’s gentle direction all the more.</p>
<p>I feel very sad knowing that Jan is no longer with us. My thoughts are with his wife, Ruth, and the rest of her family.</p>nickgillA week ago I heard the very sad news that Jan Saxl has died. Jan was my PhD supervisor at Cambridge – I spent three and a half years working under his guidance. The opportunity to work with Jan was one of the great privileges of my mathematical life, and I owe him a great debt for helping me through the difficult process of researching and writing a doctoral thesis. As well as being a fine mathematician, Jan was a warm and witty man. Now I have PhD students of my own I know that it is not always easy to know what the right thing to do is when it comes to supervision – looking back I appreciate Jan’s gentle direction all the more. I feel very sad knowing that Jan is no longer with us. My thoughts are with his wife, Ruth, and the rest of her family.An investigation of triality2020-04-23T00:00:00+00:002020-04-23T00:00:00+00:00https://nickpgill.github.io/matrices-for-O8-G2-and-3D4<script type="text/x-mathjax-config">
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<p>This post is a follow-up to <a href="https://nickpgill.github.io/matrices-for-classical-groups">my earlier post on matrices for classical groups</a>.</p>
<p>I wanted to do an investigation of the triality automorphism of the orthogonal group $O8^+(q)$. I finally got around to this today, and have written up some rough notes which are <a href="https://nickpgill.github.io/omega8.pdf">here</a>. One unexpected bonus was that I was able to write down matrices for the natural 8-dimensional representation of $G2(q)$ over $\mathbb{F}{q}$ and ${^3D4}(q)$ over $\mathbb{F}_{q^3}$.</p>
<p>Both of these families of exceptional groups lie inside $O_8^+$ groups. Their structure, when written as matrices, is surprisingly similar (surprising to me!). Although all of this is well-known to experts, I’ve not seen the matrices written down before so I was pleased that I could nut it out…</p>
<p>(By the way, I’m having trouble with underscores rendering correctly, so please do whatever’s necessary to make that middle paragraph make sense.)</p>nickgillThis post is a follow-up to my earlier post on matrices for classical groups. I wanted to do an investigation of the triality automorphism of the orthogonal group $O8^+(q)$. I finally got around to this today, and have written up some rough notes which are here. One unexpected bonus was that I was able to write down matrices for the natural 8-dimensional representation of $G2(q)$ over $\mathbb{F}{q}$ and ${^3D4}(q)$ over $\mathbb{F}_{q^3}$. Both of these families of exceptional groups lie inside $O_8^+$ groups. Their structure, when written as matrices, is surprisingly similar (surprising to me!). Although all of this is well-known to experts, I’ve not seen the matrices written down before so I was pleased that I could nut it out… (By the way, I’m having trouble with underscores rendering correctly, so please do whatever’s necessary to make that middle paragraph make sense.)Groups and Geometries 20192019-09-06T00:00:00+00:002019-09-06T00:00:00+00:00https://nickpgill.github.io/groups-and-geometries-2019<script type="text/x-mathjax-config">
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<p>I just got back from a conference at <a href="http://www.birs.ca">Banff International Research Station</a> on “Groups and Geometries”. Many thanks to Martin Liebeck, Inna Capdebosq and Bernhard Muehlherr for organising a brilliant week.</p>
<p>I gave a talk at the conference entitled “The relational complexity of a finite primitive permutation group”, which you can view by clicking <a href="http://www.birs.ca/events/2019/5-day-workshops/19w5046/videos/embed/201908261531-Gill.mp4">here</a>. The abstract of the talk was the following:</p>
<blockquote>
<p>Motivated by questions in model theory, Greg Cherlin introduced the idea of “relational
complexity”, a statistic connected to finite permutation groups. He also stated a conjecture
classifying those permutation groups with minimal relational complexity. We report on recent
progress towards a proof of this conjecture. We also make some remarks about permutation
groups with large relational complexity, and we explain how this statistic relates to others in the literature, notably base-size.</p>
<p>This work is joint with Pablo Spiga, Martin Liebeck, Francesca Dalla Volta, Francis Hunt and Bianca Lodà.”</p>
</blockquote>
<p><strong>Erratum</strong>: at minute 39 of the talk I mis-stated a theorem. The correct statement is as follows:</p>
<p><strong>Theorem</strong> (Gill, Lodà, Spiga) There exists a constant $c$ such that if $G$ acts primitively on a set $X$ of size $t$, then either</p>
<ol>
<li>$H(G,X) < c {\rm log} t$</li>
<li>$G$ is a subgroup of $S_m \wr S_r$ containing $(A_m)^r$, where the action of $S_m$ is on $k$-element subsets of ${1, …, m}$ and the wreath product has the product action of degree $t={m \choose k}^r$.</li>
</ol>nickgillI just got back from a conference at Banff International Research Station on “Groups and Geometries”. Many thanks to Martin Liebeck, Inna Capdebosq and Bernhard Muehlherr for organising a brilliant week. I gave a talk at the conference entitled “The relational complexity of a finite primitive permutation group”, which you can view by clicking here. The abstract of the talk was the following: Motivated by questions in model theory, Greg Cherlin introduced the idea of “relational complexity”, a statistic connected to finite permutation groups. He also stated a conjecture classifying those permutation groups with minimal relational complexity. We report on recent progress towards a proof of this conjecture. We also make some remarks about permutation groups with large relational complexity, and we explain how this statistic relates to others in the literature, notably base-size. This work is joint with Pablo Spiga, Martin Liebeck, Francesca Dalla Volta, Francis Hunt and Bianca Lodà.” Erratum: at minute 39 of the talk I mis-stated a theorem. The correct statement is as follows: Theorem (Gill, Lodà, Spiga) There exists a constant $c$ such that if $G$ acts primitively on a set $X$ of size $t$, then either $H(G,X) < c {\rm log} t$ $G$ is a subgroup of $S_m \wr S_r$ containing $(A_m)^r$, where the action of $S_m$ is on $k$-element subsets of ${1, …, m}$ and the wreath product has the product action of degree $t={m \choose k}^r$.Maximal tori of groups of Lie type2019-07-11T00:00:00+00:002019-07-11T00:00:00+00:00https://nickpgill.github.io/maximal-tori-of-groups-of-Lie-type<script type="text/x-mathjax-config">
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<p>I’ve been re-reading parts of Carter’s “Finite groups of Lie type” and Malle-Testerman’s “Linear algebraic groups and finite groups of Lie type” with a view to understanding the theory of maximal tori in finite groups of Lie type.</p>
<p>In this post I want to use the theory in those books to write down the orders of the maximal tori of $A_2(q)$, ${^2A_2(q)}$ and $G_2(q)$. I wanted to also do ${^2G_2(q)}$ but, so far, I haven’t managed to write things down properly for the Ree and Suzuki groups, so I’ll exclude these from what follows.</p>
<p>The general set-up is as follows: $G$ is a simple linear algebraic groups, and $F:G\to G$ is a Steinberg endomorphism, i.e. some power $F^m:G\to G$ is a Frobenius endomorphism of $G$. (Some of this theory works more generally – for $G$ connected reductive – and, in particular, this can be important when one studies centralizers inside a simple LAG’s…. But I’m not going there just now.) Now a theorem of Steinberg asserts that $G^F$, the set of fixed-points of $F$, is a finite set [MT, Theorem 21.5] – such a group is an example of a <strong>finite group of Lie type</strong>.</p>
<p>Now the Lang-Steinberg theorem asserts that the map $L:G\to G, \, g\mapsto F(g) g^{-1}$ is surjective [MT, Theorem 21.7]. This theorem then implies that $G$ contains an $F$-stable maximal torus $T$ inside an $F$-stable Borel subgroup $B$. Since $T$ is $F$-stable, $N_G(T)$ is also $F$-stable, and so $F$ naturally acts on the Weyl group $W=N_G(T)/T$ of $G$. Similarly, $F$ acts on the character group $X:=X(T)$ via</p>
\[F(\chi(t)):= \chi(F(t)) \textrm{ for } \chi \in X, t\in T.\]
<p>We will need the notion of $F$-conjugacy in $W$: if $w_1, w_2\in W$, then $w_1$ is <strong>$F$-conjugate</strong> with $w_2$ if there exists $g\in W$ such that $w_1=F(g)w_2g^{-1}$.</p>
<p>Let $\Phi\subset X$ be the root system of $G$ with positive system $\Phi^+$ with respect to $T$ and $B$. In what follows we write $X_\mathbb{R}:=X\otimes_{\mathbb{Z}}\mathbb{R}$.</p>
<p>Now the first three results of [MT, Section 22.1] imply that</p>
<ul>
<li>there exists a natural number $\delta$ such that $F^\delta=r.1$ in its action on $X$, where $r$ is some power of $p$;</li>
<li>there exists a permutation $\rho$ of $\Phi^+$ such that, for each $\alpha\in\Phi^+$, $F(\rho(\alpha)) = q_\alpha \alpha$ where $q_\alpha>1$ is a power of $p$;</li>
<li>the parameter $q_\alpha$ is constant on root lengths; moreover, either $q_\alpha=q$ or else $(G,p)\in{(B_2,2), (G_2,3), (F_4,2)}$, $\rho$ interchanges long and short roots, and $q_{long}.q_{short}=q^2$ with $q_{short}/q_{long}=p$;</li>
<li>setting $q=r^{1/\delta}$, we have that $F=q\phi$, where $q$ is the Frobenius endomorphism, and $\phi\in {\rm Aut}(X_\mathbb{R})$ is of order $\delta$ inducing $\rho^{-1}$ on $\Phi^+$;</li>
<li>$T^F=X/(F-1)/X$.</li>
</ul>
<p>It is important to note that, in principle, $q$ is a fractional power of $p$ (although, in fact, it will be integral except when $G^F$ is Ree or Suzuki). Note, too, that this set-up clearly defines the real number $q$ to be associated to our finite group of Lie type – for certain families (e.g. the unitaries), the value of $q$ follows varying conventions whereas here it is clear cut.</p>
<p>We have set-up all the necessary parameters associated with our group of Lie type. Now let’s study the maximal tori: we follow [MT, Chapter 25]. First off, we note that, since Frobenius endomorphism commute with elements of $W$ in their action on $T$, the notion of $F$-conjugacy is the same as $\phi$-conjugacy (where $F=q\phi$).</p>
<p>The following principles are important:</p>
<ul>
<li>[MT, Prop. 25.1] The $G^F$-classes of $F$-stable maximal tori of $G$ are in 1-1 correspondence with the $\phi$-conjugacy classes in $W$.</li>
<li>[MT, Exercise 30.5] The $G^F$-classes of subgroups of the form $T^F$ ($T$ an $F$-stable maximal torus of $G$) are in 1-1 correspondence with the two previous sets. (I’m slightly unsure of this… But it seems correct.)</li>
</ul>
<p>These correspondences follow from the Lang-Steinberg theorem. More precisely the first correspondence is as follows: if $gTg^{-1}$ is $F$-stable, then it corresponds to the element $w:=g^{-1}F(g)T\in N_g(T)/T=W$. We are then able to write $T_w$ for the conjugate $gTg^{-1}$. Note that $T_1$ corresponds to an $F$-stable maximal torus in an $F$-stable Borel subgroup. Now [MT, Prop. 25.3] asserts:</p>
<ul>
<li>$T_w^F\cong X/(wF-1)X$;</li>
<li>$|T_w^F|=|\det_{X\otimes \mathbb{R}}(wF-1)|=\det_{X\otimes \mathbb{R}}(q-(w\phi)^{-1})$.</li>
</ul>
<p>Specific calculations now follow. These can be confirmed using Kantor-Seress “Prime power graphs for groups of Lie type”.</p>
<hr />
<h2 id="calculations-for-a_2q">Calculations for $A_2(q)$</h2>
<p>We record the size of the maximal tori for $A_2(q)$. Note that, here and below, the isogeny class does not matter – so, in this case, these calculations are valid for ${\rm PGL}_3(q)$ and ${\rm SL}_3(q)$.</p>
<p><img src="A2.png" alt="" /></p>
<p>We use the fact that the fundamental roots of $A_2$ – labelled $\alpha$ and $\beta$ in the diagram – form a basis for $X\otimes\mathbb{R}$. With respect to this basis we have
\(q=\left(\begin{matrix} q & 0 \\ 0 & q \end{matrix}\right).\)
In this case $\phi$ is trivial, so we just need to write down $q-w^{-1}$. The possibilities are as follows:</p>
<ul>
<li>$w=1$. Then \(w=\left(\begin{matrix} 1 & 0 \\ 0 & 1\end{matrix}\right)\) and</li>
</ul>
\[\det(q-w^{-1})=\det\left(\begin{matrix} q-1 & 0 \\ 0 & q-1\end{matrix}\right) =(q-1)^2.\]
<ul>
<li>$w=Ref_\alpha$. Then \(w=\left(\begin{matrix} -1 & 1\\ 0 & 1\end{matrix}\right)\) and</li>
</ul>
\[\det(q-w^{-1})=\det\left(\begin{matrix} q+1 & -1 \\ 0 & q-1\end{matrix}\right) = q^2-1.\]
<ul>
<li>$w=Rot_{\pi/3}$. Then \(w=\left(\begin{matrix} 0 & -1\\ 1 & -1\end{matrix}\right)\) and</li>
</ul>
\[\det(q-w^{-1}=\det\left(\begin{matrix} q & 1 \\ -1 & q+1\end{matrix}\right) = q^2+q+1.\]
<hr />
<h2 id="calculations-for-2a_2q">Calculations for ${^2A_2}(q)$</h2>
<p>We record the size of the maximal tori for ${^2A_2}(q)$. The root system is as before, and we have the same value for $q$, but this time time $\phi$ is non-trivial.</p>
<p>Using the same basis as before – ${\alpha, \beta}$, we can write $\phi$ as \(\left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right)\). This is just taking $\phi$ acting on the Dynkin diagram. (Stupid comment: I’ve never cottoned on to the fact, hitherto, that $\phi$ is also an automorphism of the root system. In particular it normalizes the Weyl group which here is $W\cong D_6$. So we get $\langle W, \phi\rangle \cong D_{12}$. This is clearly true in general.)</p>
<p>So now we need to write down $q-(\phi w)^{-1}$. The possibilities for $w$ are as before:</p>
<ul>
<li>$w=1$. Then \(w=\left(\begin{matrix} 1 & 0 \\ 0 & 1\end{matrix}\right)\) and</li>
</ul>
\[\det(q-(\phi w)^{-1})=\det\left(\begin{matrix} q & 1 \\ 1 & q\end{matrix}\right) = q^2-1.\]
<ul>
<li>$w=Ref_\alpha$. Then \(w=\left(\begin{matrix} -1 & 1\\ 0 & 1\end{matrix}\right)\) and</li>
</ul>
\[\det(q-(\phi w)^{-1})=\det\left(\begin{matrix} q-1 & 1 \\ -1 & q\end{matrix}\right) = q^2-q+1.\]
<ul>
<li>$w=Ref_{\alpha+\beta}$. Then \(w=\left(\begin{matrix} 0 & -1\\ -1 & 0\end{matrix}\right)\) and</li>
</ul>
\[\det(q-(\phi w)^{-1})=\det\left(\begin{matrix} q+1 & 0 \\ 0 & q+1\end{matrix}\right) = (q+1)^2.\]
<p>Note that we need to choose different elements $w$ because the $\phi$-conjugacy classes in $W$ are different to the usual conjugacy classes.</p>
<hr />
<h2 id="calculations-for-g_2q">Calculations for $G_2(q)$</h2>
<p>Recall that $G_2(q)$ has order $q^6(q^2-1)(q^2-1)$.</p>
<p><img src="G2.png" alt="" /></p>
<p>As before, we take ${\alpha, \beta}$ as a basis for $X\otimes\mathbb{R}$, and we note that $\phi$ is trivial, and $q$ is as before. We must go through representatives for each of the conjugacy classes of $W=D_{12}$:</p>
<ul>
<li>$w=1$. Then \(w=\left(\begin{matrix} 1 & 0 \\ 0 & 1\end{matrix}\right)\) and</li>
</ul>
\[\det(q-w^{-1})=\det\left(\begin{matrix} q-1 & 0 \\ 0 & q-1\end{matrix}\right) =(q-1)^2.\]
<ul>
<li>$w=-1=Rot_{\pi/2}$. Then \(w=\left(\begin{matrix} -1 & 0\\ 0 & -1\end{matrix}\right)\) and</li>
</ul>
\[\det(q-w^{-1})=\det\left(\begin{matrix} q+1 & 0 \\ 0 & q+1\end{matrix}\right) = (q+1)^2.\]
<ul>
<li>$w=Rot_{\pi/3}$. Then \(w=\left(\begin{matrix} 1 & -3\\ 1 & -2\end{matrix}\right)\) and</li>
</ul>
\[\det(q-w^{-1})=\det\left(\begin{matrix} q-1 & -1 \\ 3 & q+2\end{matrix}\right) = q^2+q+1.\]
<ul>
<li>$w=Rot_{\pi/6}$. Then \(w=\left(\begin{matrix} 2 & -3\\ 1 & -1\end{matrix}\right)\) and</li>
</ul>
\[\det(q-w^{-1})=\det\left(\begin{matrix} q-2 & 3 \\ -1 & q+1\end{matrix}\right) = q^2-q+1.\]
<ul>
<li>$w=Ref_{3\alpha+2\beta}$. Then \(w=\left(\begin{matrix} 1 & -3\\ 0 & -1\end{matrix}\right)\) and</li>
</ul>
\[\det(q-w^{-1})=\det\left(\begin{matrix} q-1 & 3 \\ 0 & q+1\end{matrix}\right) = q^2-1.\]
<p>This yields all of the maximal tori that are listed in Kantor-Seress. However note that there are two conjugacy classes of reflections in $D_{12}$ – here they correspond to reflections in long and short roots – and so we obtain another example:</p>
<ul>
<li>$w=Ref_{\alpha+\beta}$. Then \(w=\left(\begin{matrix} 2 & -3\\ 1 & -2\end{matrix}\right)\) and</li>
</ul>
\[\det(q-w^{-1})=\det\left(\begin{matrix} q-2 & 3 \\ -1 & q+2\end{matrix}\right) = q^2-1.\]
<p>Thus it appears that there are two conjugacy classes of maximal torus of order $q^2-1$ – I guess one occurs as a split torus in a Levi factor ${\rm GL}_2(q)=\langle U_\alpha, U_{-\alpha}\rangle$, while the other occurs in ${\rm GL}_2(q)=\langle U_\beta, U_{-\beta}\rangle$. I reconcile this to Kantor-Seress by noting that they do not necessarily claim to list all conjugacy classes of tori, although in some places they do note that there is more than one conjugacy class of a certain order.</p>nickgillI’ve been re-reading parts of Carter’s “Finite groups of Lie type” and Malle-Testerman’s “Linear algebraic groups and finite groups of Lie type” with a view to understanding the theory of maximal tori in finite groups of Lie type. In this post I want to use the theory in those books to write down the orders of the maximal tori of $A_2(q)$, ${^2A_2(q)}$ and $G_2(q)$. I wanted to also do ${^2G_2(q)}$ but, so far, I haven’t managed to write things down properly for the Ree and Suzuki groups, so I’ll exclude these from what follows. The general set-up is as follows: $G$ is a simple linear algebraic groups, and $F:G\to G$ is a Steinberg endomorphism, i.e. some power $F^m:G\to G$ is a Frobenius endomorphism of $G$. (Some of this theory works more generally – for $G$ connected reductive – and, in particular, this can be important when one studies centralizers inside a simple LAG’s…. But I’m not going there just now.) Now a theorem of Steinberg asserts that $G^F$, the set of fixed-points of $F$, is a finite set [MT, Theorem 21.5] – such a group is an example of a finite group of Lie type. Now the Lang-Steinberg theorem asserts that the map $L:G\to G, \, g\mapsto F(g) g^{-1}$ is surjective [MT, Theorem 21.7]. This theorem then implies that $G$ contains an $F$-stable maximal torus $T$ inside an $F$-stable Borel subgroup $B$. Since $T$ is $F$-stable, $N_G(T)$ is also $F$-stable, and so $F$ naturally acts on the Weyl group $W=N_G(T)/T$ of $G$. Similarly, $F$ acts on the character group $X:=X(T)$ via \[F(\chi(t)):= \chi(F(t)) \textrm{ for } \chi \in X, t\in T.\] We will need the notion of $F$-conjugacy in $W$: if $w_1, w_2\in W$, then $w_1$ is $F$-conjugate with $w_2$ if there exists $g\in W$ such that $w_1=F(g)w_2g^{-1}$. Let $\Phi\subset X$ be the root system of $G$ with positive system $\Phi^+$ with respect to $T$ and $B$. In what follows we write $X_\mathbb{R}:=X\otimes_{\mathbb{Z}}\mathbb{R}$. Now the first three results of [MT, Section 22.1] imply that there exists a natural number $\delta$ such that $F^\delta=r.1$ in its action on $X$, where $r$ is some power of $p$; there exists a permutation $\rho$ of $\Phi^+$ such that, for each $\alpha\in\Phi^+$, $F(\rho(\alpha)) = q_\alpha \alpha$ where $q_\alpha>1$ is a power of $p$; the parameter $q_\alpha$ is constant on root lengths; moreover, either $q_\alpha=q$ or else $(G,p)\in{(B_2,2), (G_2,3), (F_4,2)}$, $\rho$ interchanges long and short roots, and $q_{long}.q_{short}=q^2$ with $q_{short}/q_{long}=p$; setting $q=r^{1/\delta}$, we have that $F=q\phi$, where $q$ is the Frobenius endomorphism, and $\phi\in {\rm Aut}(X_\mathbb{R})$ is of order $\delta$ inducing $\rho^{-1}$ on $\Phi^+$; $T^F=X/(F-1)/X$. It is important to note that, in principle, $q$ is a fractional power of $p$ (although, in fact, it will be integral except when $G^F$ is Ree or Suzuki). Note, too, that this set-up clearly defines the real number $q$ to be associated to our finite group of Lie type – for certain families (e.g. the unitaries), the value of $q$ follows varying conventions whereas here it is clear cut. We have set-up all the necessary parameters associated with our group of Lie type. Now let’s study the maximal tori: we follow [MT, Chapter 25]. First off, we note that, since Frobenius endomorphism commute with elements of $W$ in their action on $T$, the notion of $F$-conjugacy is the same as $\phi$-conjugacy (where $F=q\phi$). The following principles are important: [MT, Prop. 25.1] The $G^F$-classes of $F$-stable maximal tori of $G$ are in 1-1 correspondence with the $\phi$-conjugacy classes in $W$. [MT, Exercise 30.5] The $G^F$-classes of subgroups of the form $T^F$ ($T$ an $F$-stable maximal torus of $G$) are in 1-1 correspondence with the two previous sets. (I’m slightly unsure of this… But it seems correct.) These correspondences follow from the Lang-Steinberg theorem. More precisely the first correspondence is as follows: if $gTg^{-1}$ is $F$-stable, then it corresponds to the element $w:=g^{-1}F(g)T\in N_g(T)/T=W$. We are then able to write $T_w$ for the conjugate $gTg^{-1}$. Note that $T_1$ corresponds to an $F$-stable maximal torus in an $F$-stable Borel subgroup. Now [MT, Prop. 25.3] asserts: $T_w^F\cong X/(wF-1)X$; $|T_w^F|=|\det_{X\otimes \mathbb{R}}(wF-1)|=\det_{X\otimes \mathbb{R}}(q-(w\phi)^{-1})$. Specific calculations now follow. These can be confirmed using Kantor-Seress “Prime power graphs for groups of Lie type”. Calculations for $A_2(q)$ We record the size of the maximal tori for $A_2(q)$. Note that, here and below, the isogeny class does not matter – so, in this case, these calculations are valid for ${\rm PGL}_3(q)$ and ${\rm SL}_3(q)$. We use the fact that the fundamental roots of $A_2$ – labelled $\alpha$ and $\beta$ in the diagram – form a basis for $X\otimes\mathbb{R}$. With respect to this basis we have \(q=\left(\begin{matrix} q & 0 \\ 0 & q \end{matrix}\right).\) In this case $\phi$ is trivial, so we just need to write down $q-w^{-1}$. The possibilities are as follows: $w=1$. Then \(w=\left(\begin{matrix} 1 & 0 \\ 0 & 1\end{matrix}\right)\) and \[\det(q-w^{-1})=\det\left(\begin{matrix} q-1 & 0 \\ 0 & q-1\end{matrix}\right) =(q-1)^2.\] $w=Ref_\alpha$. Then \(w=\left(\begin{matrix} -1 & 1\\ 0 & 1\end{matrix}\right)\) and \[\det(q-w^{-1})=\det\left(\begin{matrix} q+1 & -1 \\ 0 & q-1\end{matrix}\right) = q^2-1.\] $w=Rot_{\pi/3}$. Then \(w=\left(\begin{matrix} 0 & -1\\ 1 & -1\end{matrix}\right)\) and \[\det(q-w^{-1}=\det\left(\begin{matrix} q & 1 \\ -1 & q+1\end{matrix}\right) = q^2+q+1.\] Calculations for ${^2A_2}(q)$ We record the size of the maximal tori for ${^2A_2}(q)$. The root system is as before, and we have the same value for $q$, but this time time $\phi$ is non-trivial. Using the same basis as before – ${\alpha, \beta}$, we can write $\phi$ as \(\left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right)\). This is just taking $\phi$ acting on the Dynkin diagram. (Stupid comment: I’ve never cottoned on to the fact, hitherto, that $\phi$ is also an automorphism of the root system. In particular it normalizes the Weyl group which here is $W\cong D_6$. So we get $\langle W, \phi\rangle \cong D_{12}$. This is clearly true in general.) So now we need to write down $q-(\phi w)^{-1}$. The possibilities for $w$ are as before: $w=1$. Then \(w=\left(\begin{matrix} 1 & 0 \\ 0 & 1\end{matrix}\right)\) and \[\det(q-(\phi w)^{-1})=\det\left(\begin{matrix} q & 1 \\ 1 & q\end{matrix}\right) = q^2-1.\] $w=Ref_\alpha$. Then \(w=\left(\begin{matrix} -1 & 1\\ 0 & 1\end{matrix}\right)\) and \[\det(q-(\phi w)^{-1})=\det\left(\begin{matrix} q-1 & 1 \\ -1 & q\end{matrix}\right) = q^2-q+1.\] $w=Ref_{\alpha+\beta}$. Then \(w=\left(\begin{matrix} 0 & -1\\ -1 & 0\end{matrix}\right)\) and \[\det(q-(\phi w)^{-1})=\det\left(\begin{matrix} q+1 & 0 \\ 0 & q+1\end{matrix}\right) = (q+1)^2.\] Note that we need to choose different elements $w$ because the $\phi$-conjugacy classes in $W$ are different to the usual conjugacy classes. Calculations for $G_2(q)$ Recall that $G_2(q)$ has order $q^6(q^2-1)(q^2-1)$. As before, we take ${\alpha, \beta}$ as a basis for $X\otimes\mathbb{R}$, and we note that $\phi$ is trivial, and $q$ is as before. We must go through representatives for each of the conjugacy classes of $W=D_{12}$: $w=1$. Then \(w=\left(\begin{matrix} 1 & 0 \\ 0 & 1\end{matrix}\right)\) and \[\det(q-w^{-1})=\det\left(\begin{matrix} q-1 & 0 \\ 0 & q-1\end{matrix}\right) =(q-1)^2.\] $w=-1=Rot_{\pi/2}$. Then \(w=\left(\begin{matrix} -1 & 0\\ 0 & -1\end{matrix}\right)\) and \[\det(q-w^{-1})=\det\left(\begin{matrix} q+1 & 0 \\ 0 & q+1\end{matrix}\right) = (q+1)^2.\] $w=Rot_{\pi/3}$. Then \(w=\left(\begin{matrix} 1 & -3\\ 1 & -2\end{matrix}\right)\) and \[\det(q-w^{-1})=\det\left(\begin{matrix} q-1 & -1 \\ 3 & q+2\end{matrix}\right) = q^2+q+1.\] $w=Rot_{\pi/6}$. Then \(w=\left(\begin{matrix} 2 & -3\\ 1 & -1\end{matrix}\right)\) and \[\det(q-w^{-1})=\det\left(\begin{matrix} q-2 & 3 \\ -1 & q+1\end{matrix}\right) = q^2-q+1.\] $w=Ref_{3\alpha+2\beta}$. Then \(w=\left(\begin{matrix} 1 & -3\\ 0 & -1\end{matrix}\right)\) and \[\det(q-w^{-1})=\det\left(\begin{matrix} q-1 & 3 \\ 0 & q+1\end{matrix}\right) = q^2-1.\] This yields all of the maximal tori that are listed in Kantor-Seress. However note that there are two conjugacy classes of reflections in $D_{12}$ – here they correspond to reflections in long and short roots – and so we obtain another example: $w=Ref_{\alpha+\beta}$. Then \(w=\left(\begin{matrix} 2 & -3\\ 1 & -2\end{matrix}\right)\) and \[\det(q-w^{-1})=\det\left(\begin{matrix} q-2 & 3 \\ -1 & q+2\end{matrix}\right) = q^2-1.\] Thus it appears that there are two conjugacy classes of maximal torus of order $q^2-1$ – I guess one occurs as a split torus in a Levi factor ${\rm GL}_2(q)=\langle U_\alpha, U_{-\alpha}\rangle$, while the other occurs in ${\rm GL}_2(q)=\langle U_\beta, U_{-\beta}\rangle$. I reconcile this to Kantor-Seress by noting that they do not necessarily claim to list all conjugacy classes of tori, although in some places they do note that there is more than one conjugacy class of a certain order.Tuna Altinel2019-06-06T00:00:00+00:002019-06-06T00:00:00+00:00https://nickpgill.github.io/tuna-altinel<p>I was recently alerted to a <a href="https://euro-math-soc.eu/news/19/05/14/ems-statement-arrest-prof-tuna-altinel">European Maths Society statement on the arrest of Professor Tuna Altinel</a>. I reproduce that statement below. The situation sounds bad, and I so I am posting here to alert members of the mathematical community to what is going on. If you wish to show solidarity with Professor Altinel, then you can do so by signing the petition <a href="http://math.univ-lyon1.fr/SoutienTunaAltinel/?lang=en">here</a>.</p>
<h3 id="ems-statement-on-the-arrest-of-prof-tuna-altinel">EMS Statement on the Arrest of Prof Tuna Altinel</h3>
<blockquote>
<p>Last week the mathematician Tuna Altinel, member of the European Mathematical Society and professor at the Université Lyon 1 in France, was arrested in Turkey after he had his passport extracted by the police. Tuna Altinel was one of the signatories of the peace petition supported by more than 2000 scientists and intellectuals against military actions towards civilians.</p>
</blockquote>
<blockquote>
<p>The European Mathematical Society condemns this violation of Prof Altinel’s human rights and demands that he is immediately released and allowed to return to France to resume his teaching and research.</p>
</blockquote>nickgillI was recently alerted to a European Maths Society statement on the arrest of Professor Tuna Altinel. I reproduce that statement below. The situation sounds bad, and I so I am posting here to alert members of the mathematical community to what is going on. If you wish to show solidarity with Professor Altinel, then you can do so by signing the petition here. EMS Statement on the Arrest of Prof Tuna Altinel Last week the mathematician Tuna Altinel, member of the European Mathematical Society and professor at the Université Lyon 1 in France, was arrested in Turkey after he had his passport extracted by the police. Tuna Altinel was one of the signatories of the peace petition supported by more than 2000 scientists and intellectuals against military actions towards civilians. The European Mathematical Society condemns this violation of Prof Altinel’s human rights and demands that he is immediately released and allowed to return to France to resume his teaching and research.A Rodgers-Saxl type conjecture for characters2019-04-05T00:00:00+00:002019-04-05T00:00:00+00:00https://nickpgill.github.io/a-rodgers-saxl-conjecture-for-characters<script type="text/x-mathjax-config">
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<p>This is a follow-up to my <a href="https://nickpgill.github.io/a-rodgers-saxl-theorem">previous post on a generalization of a theorem of Rodgers and Saxl</a>.</p>
<p>The story starts with a beautiful talk I heard by Martin Liebeck in which he outlined a result due to him, Shalev & Tiep:</p>
<p><strong>Theorem</strong>: Let $f$ be a faithful irreducible of $Sym(n)$. Then $f^{*4n}$ (i.e.tensor product $4n$ times) contains every irreducible as a constituent.</p>
<p>This theorem can be interpreted as giving an upper bound on the diameter of the <a href="https://en.wikipedia.org/wiki/McKay_graph">McKay graphs</a> of the symmetric group. I won’t pursue this point of view, but it was within this context that the LST-team were working when they proved the theorem. I’m just stating the symmetric group version of their result – they had more general statements for finite (almost) simple groups.</p>
<p>Notice that the LPT-bound is pretty much as good as one could hope for: if $f^{*c}$ is to contain every irreducible as a constituent (for some positive integer $c$), then one needs $(\dim(f))^c > \sum \dim(f_i)$ where the sum on the right hand side ranges over all irreducibles of $Sym(n)$. Now the theory of Frobenius-Schur indicators tells us that, since all complex reps of $Sym(n)$ are defined over the reals, then $\sum \dim(f_i)$ is equal to 1+ the number of elements of order $2$. Writing $I_n$ for this latter quantity, <a href="https://projecteuclid.org/download/pdf_1/euclid.bams/1183553478">a result of Moser and Wyman</a> asserts that</p>
\[I_n \sim\frac{1}{\sqrt{2}} n^{n/2} \exp(-n/2-1/4+\sqrt{n}).\]
<p>Now, using the fact that there exists an irreducible of dimension $n-1$, we obtain that</p>
\[c>= \log (\sum \dim(f_i))/ \log(\dim(f) = \log (I_n+1)/ \log (n-1),\]
<p>and we conclude that $c$ must be at least linear in $n$.</p>
<p>A well-known heuristic in finite group theory says that whenever one proves statements about (ordinary) characters, there is probably a statement about conjugacy classes lurking nearby (and vice versa). This heuristic sounds very wooly, but it can be made rigorous in very many different contexts, and in very many different ways.</p>
<p>Sure enough, there is a “conjugacy class version” of the theorem above – it was proved by Liebeck and Shalev:</p>
<p><strong>Theorem</strong>: There exists a constant $d$ such that if $C$ is a conjugacy class of $G$, a finite non-abelian simple group and if
\(k >= d \log|G|/ \log|C|,\)
then $C^k = G$.</p>
<p>Here we are taking products of conjugacy classes instead of tensor products of characters. But, again, the result is of the same kind – it says that by taking a product $d$ times, then you will obtain all conjugacy classes, and $d$ is as small as one could possibly ask for, up to a multiplicative constant.</p>
<p>So, now, recall that in the previous post, I made the following conjecture:</p>
<p><strong>Conjecture</strong>: There exists a constant $c$ such that if $G$ is a finite simple group, and $S_1,\dots, S_k$ are subsets of $G$ satisfying
$\Pi_{i=1}^k|S_i|\geq|G|^c$, then there exist elements $g_1,\dots, g_k$ such that $G=(S_1)^{g_1}\cdots (S_k)^{g_k}$.</p>
<p>A special case of this conjecture occurs when our sets $S_1,\dots, S_k$ are conjugacy classes of $G$. In this case, we obtain the following statement:</p>
<p><strong>Conjecture</strong>: There exists a constant $c$ such that if $G$ is a finite simple group, and $C_1,\dots, C_k$ are conjugacy classes of $G$ satisfying
$\Pi_{i=1}^k|S_i|\geq|G|^c$, then $G=C_1\cdot C_2\cdots C_k$.</p>
<p>I don’t know how to prove this theorem, but it’s possible that it’s not out of reach. The Rodgers–Saxl theorem that started all this off implies that the conjecture is true for the family $PSL(n,q)$ with the constant $c=12$. The theorem I proved with Pyber and Szabo implies it for groups of Lie type of bounded rank, so one is left with (some of) the classicals of unbounded rank, and the alternating groups.</p>
<p>But back to chacters. What would be the character–theoretic version of the previous two conjectures? The first has, if I recall correctly, been stated by the LST-team:</p>
<p><strong>Conjecture</strong> There exists $C>0$ such that if $\chi$ is a non-trivial character of a finite simple group $G$ and if
\(c>C \log(\textrm{sum of dimensions of all irreducibles of }G)/ \log(\textrm{dimension of }\chi),\)
then $\chi^{*c}$ contains every irreducible of $G$ as a constituent.</p>
<p>The Rodgers-Saxl analogue of this would be:</p>
<p><strong>Conjecture</strong> There exists $C>0$ such that if $\chi_1,…, \chi_t$ are non-trivial characters of a finite simple group $G$ and if</p>
\[\dim(\chi_1)*\dim(\chi_2)*...*\dim(\chi_t) > (\textrm{sum of dimensions of all irreducibles of }G)^C\]
<p>then $\chi_{1}*\cdots *\chi_{t}$ contains every irreducible of $G$ as a constituent.</p>
<p>Let’s think about whether proving something like this might be possible just for the special case of $G=Sym(n)$ (OK, it’s not simple, but almost).</p>
<p>First let me note that LST prove their result by showing that if $f$ is a faithful character of $Sym(n)$, then $f * f$ or $f *f *f *f$ always contains $\chi_{1,n-1}$, from which the result follows (one just calculates how many tensor products of $\chi_{1,n-1}$ one needs to obtain all irreducibles as constituents).</p>
<p>My go-to man for symmetric group rep theory is Mark Wildon. I dropped him an email with the following question.</p>
<p><strong>Question</strong>: Does there exist a positive integer $N$ such that if $k> N$ and $f_1, …, f_k$ are irreducible characters of Sym(n), then the tensor product of $f_1,…, f_k$ (in that order) contains an irreducible constituent isomorphic to $\chi_{1,n-1}$?</p>
<p>Mark was immediately able to shed light. This question can be answered affirmatively. Indeed the following argument (which is Mark’s) shows that something much stronger is true:</p>
<p>Let $k$ be a field and let $G$ be a finite group. On page 45 of Alperin, <em>Local Representation Theory</em>, it’s shown that if $V$ is a faithful $k$G-module then there exists $N$ such that the $N$-fold tensor product $V \otimes … \otimes V$ contains a free submodule. Since $V \otimes F \cong F \oplus … \oplus F$ (with $\dim(V)$ summands) for any free $kG$-module $F$, it follows that any product with more factors (which may be arbitrary kG-modules) also contains a free submodule.</p>
<p>In the language of characters: if $\chi$ is the character of $V$, and $\psi$ is any other character, then any character $\chi^N \times \psi$ contains the regular character.</p>
<p><strong>Corollary</strong>: there exists $M$ such that any product of any $M$ faithful irreducible characters of $Sym(n)$ contains the regular character as a constituent.</p>
<p><strong>Proof</strong>: let $P$ be the number of faithful irreducible characters of $Sym(n)$. (So $P$ is 2 less than the number of partitions of $n$, unless $n = 4$.) For each faithful character, let $N(\chi)$ be the $N$ given by Alperin’s result, and let $N = \max_\chi N(\chi)$. Take $M = NP$. Then in any product of $M$ faithful characters, some character appears at least $N$ times, and so the product contains the regular character. QED</p>
<p>That’s a brilliant start, but it doesn’t give us any information about what $M$ can be. The same argument as I gave at the top of this post implies that $M$ must be bounded below by some linear function in $n$: so, then, it it possible that one can choose $M$ to be linear in $n$?</p>
<p>Here’s what Mark had to say on the subject:</p>
<blockquote>
<p><em>Some experiments in MAGMA suggest rather intriguingly that one can take
$M = n - 1$ for $Sym(n)$. This bound is sharp for $n = 3,\dots, 10$.</em></p>
</blockquote>
<p>Is this enough evidence to make a conjecture? Hell, yeah!</p>
<p><strong>Conjecture</strong>: Suppose that $f_1,\dots, f_{n-1}$ are faithful irreducible characters of $Sym(n)$. Then $f_1* \cdots* f_{n-1}$ contains the regular character as a constituent.</p>
<p>I think of this as a “Rodgers-Saxl type conjecture for characters”. Now, the challenge is to turn it into a theorem….</p>nickgillThis is a follow-up to my previous post on a generalization of a theorem of Rodgers and Saxl. The story starts with a beautiful talk I heard by Martin Liebeck in which he outlined a result due to him, Shalev & Tiep: Theorem: Let $f$ be a faithful irreducible of $Sym(n)$. Then $f^{*4n}$ (i.e.tensor product $4n$ times) contains every irreducible as a constituent. This theorem can be interpreted as giving an upper bound on the diameter of the McKay graphs of the symmetric group. I won’t pursue this point of view, but it was within this context that the LST-team were working when they proved the theorem. I’m just stating the symmetric group version of their result – they had more general statements for finite (almost) simple groups. Notice that the LPT-bound is pretty much as good as one could hope for: if $f^{*c}$ is to contain every irreducible as a constituent (for some positive integer $c$), then one needs $(\dim(f))^c > \sum \dim(f_i)$ where the sum on the right hand side ranges over all irreducibles of $Sym(n)$. Now the theory of Frobenius-Schur indicators tells us that, since all complex reps of $Sym(n)$ are defined over the reals, then $\sum \dim(f_i)$ is equal to 1+ the number of elements of order $2$. Writing $I_n$ for this latter quantity, a result of Moser and Wyman asserts that \[I_n \sim\frac{1}{\sqrt{2}} n^{n/2} \exp(-n/2-1/4+\sqrt{n}).\] Now, using the fact that there exists an irreducible of dimension $n-1$, we obtain that \[c>= \log (\sum \dim(f_i))/ \log(\dim(f) = \log (I_n+1)/ \log (n-1),\] and we conclude that $c$ must be at least linear in $n$. A well-known heuristic in finite group theory says that whenever one proves statements about (ordinary) characters, there is probably a statement about conjugacy classes lurking nearby (and vice versa). This heuristic sounds very wooly, but it can be made rigorous in very many different contexts, and in very many different ways. Sure enough, there is a “conjugacy class version” of the theorem above – it was proved by Liebeck and Shalev: Theorem: There exists a constant $d$ such that if $C$ is a conjugacy class of $G$, a finite non-abelian simple group and if \(k >= d \log|G|/ \log|C|,\) then $C^k = G$. Here we are taking products of conjugacy classes instead of tensor products of characters. But, again, the result is of the same kind – it says that by taking a product $d$ times, then you will obtain all conjugacy classes, and $d$ is as small as one could possibly ask for, up to a multiplicative constant. So, now, recall that in the previous post, I made the following conjecture: Conjecture: There exists a constant $c$ such that if $G$ is a finite simple group, and $S_1,\dots, S_k$ are subsets of $G$ satisfying $\Pi_{i=1}^k|S_i|\geq|G|^c$, then there exist elements $g_1,\dots, g_k$ such that $G=(S_1)^{g_1}\cdots (S_k)^{g_k}$. A special case of this conjecture occurs when our sets $S_1,\dots, S_k$ are conjugacy classes of $G$. In this case, we obtain the following statement: Conjecture: There exists a constant $c$ such that if $G$ is a finite simple group, and $C_1,\dots, C_k$ are conjugacy classes of $G$ satisfying $\Pi_{i=1}^k|S_i|\geq|G|^c$, then $G=C_1\cdot C_2\cdots C_k$. I don’t know how to prove this theorem, but it’s possible that it’s not out of reach. The Rodgers–Saxl theorem that started all this off implies that the conjecture is true for the family $PSL(n,q)$ with the constant $c=12$. The theorem I proved with Pyber and Szabo implies it for groups of Lie type of bounded rank, so one is left with (some of) the classicals of unbounded rank, and the alternating groups. But back to chacters. What would be the character–theoretic version of the previous two conjectures? The first has, if I recall correctly, been stated by the LST-team: Conjecture There exists $C>0$ such that if $\chi$ is a non-trivial character of a finite simple group $G$ and if \(c>C \log(\textrm{sum of dimensions of all irreducibles of }G)/ \log(\textrm{dimension of }\chi),\) then $\chi^{*c}$ contains every irreducible of $G$ as a constituent. The Rodgers-Saxl analogue of this would be: Conjecture There exists $C>0$ such that if $\chi_1,…, \chi_t$ are non-trivial characters of a finite simple group $G$ and if \[\dim(\chi_1)*\dim(\chi_2)*...*\dim(\chi_t) > (\textrm{sum of dimensions of all irreducibles of }G)^C\] then $\chi_{1}*\cdots *\chi_{t}$ contains every irreducible of $G$ as a constituent. Let’s think about whether proving something like this might be possible just for the special case of $G=Sym(n)$ (OK, it’s not simple, but almost). First let me note that LST prove their result by showing that if $f$ is a faithful character of $Sym(n)$, then $f * f$ or $f *f *f *f$ always contains $\chi_{1,n-1}$, from which the result follows (one just calculates how many tensor products of $\chi_{1,n-1}$ one needs to obtain all irreducibles as constituents). My go-to man for symmetric group rep theory is Mark Wildon. I dropped him an email with the following question. Question: Does there exist a positive integer $N$ such that if $k> N$ and $f_1, …, f_k$ are irreducible characters of Sym(n), then the tensor product of $f_1,…, f_k$ (in that order) contains an irreducible constituent isomorphic to $\chi_{1,n-1}$? Mark was immediately able to shed light. This question can be answered affirmatively. Indeed the following argument (which is Mark’s) shows that something much stronger is true: Let $k$ be a field and let $G$ be a finite group. On page 45 of Alperin, Local Representation Theory, it’s shown that if $V$ is a faithful $k$G-module then there exists $N$ such that the $N$-fold tensor product $V \otimes … \otimes V$ contains a free submodule. Since $V \otimes F \cong F \oplus … \oplus F$ (with $\dim(V)$ summands) for any free $kG$-module $F$, it follows that any product with more factors (which may be arbitrary kG-modules) also contains a free submodule. In the language of characters: if $\chi$ is the character of $V$, and $\psi$ is any other character, then any character $\chi^N \times \psi$ contains the regular character. Corollary: there exists $M$ such that any product of any $M$ faithful irreducible characters of $Sym(n)$ contains the regular character as a constituent. Proof: let $P$ be the number of faithful irreducible characters of $Sym(n)$. (So $P$ is 2 less than the number of partitions of $n$, unless $n = 4$.) For each faithful character, let $N(\chi)$ be the $N$ given by Alperin’s result, and let $N = \max_\chi N(\chi)$. Take $M = NP$. Then in any product of $M$ faithful characters, some character appears at least $N$ times, and so the product contains the regular character. QED That’s a brilliant start, but it doesn’t give us any information about what $M$ can be. The same argument as I gave at the top of this post implies that $M$ must be bounded below by some linear function in $n$: so, then, it it possible that one can choose $M$ to be linear in $n$? Here’s what Mark had to say on the subject: Some experiments in MAGMA suggest rather intriguingly that one can take $M = n - 1$ for $Sym(n)$. This bound is sharp for $n = 3,\dots, 10$. Is this enough evidence to make a conjecture? Hell, yeah! Conjecture: Suppose that $f_1,\dots, f_{n-1}$ are faithful irreducible characters of $Sym(n)$. Then $f_1* \cdots* f_{n-1}$ contains the regular character as a constituent. I think of this as a “Rodgers-Saxl type conjecture for characters”. Now, the challenge is to turn it into a theorem….A Rodgers-Saxl type theorem2019-03-29T00:00:00+00:002019-03-29T00:00:00+00:00https://nickpgill.github.io/a-rodgers-saxl-theorem<script type="text/x-mathjax-config">
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<p>Pyber, Szabó and I have recently completed a paper entitled <em>A generalization of a theorem of Rodgers and Saxl for simple groups of bounded rank</em>. A copy of the paper can be found <a href="https://arxiv.org/abs/1901.09255">on the arXiv</a>.</p>
<p>Our result was inspired by a result of Rodgers and Saxl which appeared in <em>Communications of Algebra</em> in 2003:</p>
<p><strong>Theorem</strong>: Suppose that $C_1,\dots, C_k$ are conjugacy classes in $SL_n(q)$ such that
$\Pi_{i=1}^k|C_i|\geq|SL_n(q)|^{12}$. Then
$\Pi_{i=1}^kC_i=SL_n(q)$.</p>
<p>Our result has a similar flavour.</p>
<p><strong>Theorem</strong>: Let $G=G_r(q)$ be a finite simple group of Lie type of rank $r$. There exists $c=f(r)$ such that if $S_1,\dots, S_k$ are subsets of $G$ satisfying
$\Pi_{i=1}^k|S_i|\geq|G|^c$, then there exist elements $g_1,\dots, g_k$ such that $G=(S_1)^{g_1}\cdots (S_k)^{g_k}$.</p>
<p>Our theorem differs to that of Rodgers and Saxl in three important respects, two good, one not so good: First, our result pertains to all finite simple groups $G$ of Lie type. Second, our result does not just pertain to conjugacy classes, but to subsets of the group, provided we are free to take conjugates.</p>
<p>The third difference is a weak point: our result replaces the constant ``12’’ in their thereom with an unspecified constant that depends on the rank of the group $G$. We conjecture that we should be able to do better, and not just for finite simple groups of Lie type, but for alternating groups as well:</p>
<p><strong>Conjecture</strong>: Let $G$ be a finite simple group. There exists $c$ such that if $S_1,\dots, S_k$ are subsets of $G$ satisfying
$\Pi_{i=1}^k|S_i|\geq|G|^c$, then there exist elements $g_1,\dots, g_k$ such that $G=(S_1)^{g_1}\cdots (S_k)^{g_k}$.</p>
<p>This conjecture seems hard! Our theorem has a rank-dependency because it makes use of the “Product Theorem” which was proved, independently by Pyber-Szabó and by Breuillard-Green-Tao. To prove the conjecture we would need to replace the Product Theorem in our argument with, um, something else… But what?!</p>
<p>One last remark: there is a fourth sense in which our theorem differs to that of Rodgers and Saxl – we are interested in finite simple groups, while they consider $SL_n(q)$ which is, in general, only quasi-simple.</p>
<p>It turns out that the distinction here is not significant: It is not hard to show that our theorem is true if and only if the analogous statement is true for quasi-simple groups (provided you require that the sets $S_i$ do not intersect the centre of $G$)… And the same is true of the conjecture stated above. So the stated conjecture would be a generalization of both our result and that of Rodgers and Saxl, albeit we don’t specify the value of $c$ as Rodgers and Saxl did.</p>
<p>Let me finish this post by thanking my two co-authors, Laci and Bandi. I have worked with these two guys on a previous paper, and they are both brilliant and generous with their many ideas. I hope to have the privilege of working with them more in the future.</p>nickgillPyber, Szabó and I have recently completed a paper entitled A generalization of a theorem of Rodgers and Saxl for simple groups of bounded rank. A copy of the paper can be found on the arXiv. Our result was inspired by a result of Rodgers and Saxl which appeared in Communications of Algebra in 2003: Theorem: Suppose that $C_1,\dots, C_k$ are conjugacy classes in $SL_n(q)$ such that $\Pi_{i=1}^k|C_i|\geq|SL_n(q)|^{12}$. Then $\Pi_{i=1}^kC_i=SL_n(q)$. Our result has a similar flavour. Theorem: Let $G=G_r(q)$ be a finite simple group of Lie type of rank $r$. There exists $c=f(r)$ such that if $S_1,\dots, S_k$ are subsets of $G$ satisfying $\Pi_{i=1}^k|S_i|\geq|G|^c$, then there exist elements $g_1,\dots, g_k$ such that $G=(S_1)^{g_1}\cdots (S_k)^{g_k}$. Our theorem differs to that of Rodgers and Saxl in three important respects, two good, one not so good: First, our result pertains to all finite simple groups $G$ of Lie type. Second, our result does not just pertain to conjugacy classes, but to subsets of the group, provided we are free to take conjugates. The third difference is a weak point: our result replaces the constant ``12’’ in their thereom with an unspecified constant that depends on the rank of the group $G$. We conjecture that we should be able to do better, and not just for finite simple groups of Lie type, but for alternating groups as well: Conjecture: Let $G$ be a finite simple group. There exists $c$ such that if $S_1,\dots, S_k$ are subsets of $G$ satisfying $\Pi_{i=1}^k|S_i|\geq|G|^c$, then there exist elements $g_1,\dots, g_k$ such that $G=(S_1)^{g_1}\cdots (S_k)^{g_k}$. This conjecture seems hard! Our theorem has a rank-dependency because it makes use of the “Product Theorem” which was proved, independently by Pyber-Szabó and by Breuillard-Green-Tao. To prove the conjecture we would need to replace the Product Theorem in our argument with, um, something else… But what?! One last remark: there is a fourth sense in which our theorem differs to that of Rodgers and Saxl – we are interested in finite simple groups, while they consider $SL_n(q)$ which is, in general, only quasi-simple. It turns out that the distinction here is not significant: It is not hard to show that our theorem is true if and only if the analogous statement is true for quasi-simple groups (provided you require that the sets $S_i$ do not intersect the centre of $G$)… And the same is true of the conjecture stated above. So the stated conjecture would be a generalization of both our result and that of Rodgers and Saxl, albeit we don’t specify the value of $c$ as Rodgers and Saxl did. Let me finish this post by thanking my two co-authors, Laci and Bandi. I have worked with these two guys on a previous paper, and they are both brilliant and generous with their many ideas. I hope to have the privilege of working with them more in the future.Matrices for classical groups2019-01-01T00:00:00+00:002019-01-01T00:00:00+00:00https://nickpgill.github.io/matrices-for-classical-groups<script type="text/x-mathjax-config">
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<p>A finite classical group is best thought of as a group of linear operators on some vector space defined over a finite field. Which means, of course, that I can take a basis for this vector space and then represent the elements of my group as matrices.</p>
<p>However, I have almost <strong>never</strong> seen anyone do this in the literature. “Taking a basis” is thought of as a rather crude thing to do when doing linear algebra – one generally tries to write general arguments that do not refer to any particular basis. However, speaking for myself, when I’m trying to work out what the hell is going on inside a finite matrix group, I often end up trying to write down individual elements as matrices… And then hiding all this when it comes to writing up the paper!</p>
<p>This means that I have never publicly written down any of these calculations, despite using many of them over and over again. So this page is designed to be a little repository for me to note down interesting observations about such calculations… And perhaps they’ll be useful for someone else some time…</p>
<h2 id="working-with-omega_4q">Working with $\Omega_4^+(q)$</h2>
<p>This family of groups has some weird behaviour, especially when $q$ is even. For instance, let us write $\mathcal{U}$ for the set of maximal totally singular subspaces in a formed space of type $O+$ and dimension $2m$. If $q$ is even, then, provided $(m,q)\neq (2,2)$, we can define $\Omega_{2m}^+(q)$ to be the group inducing odd permutations on $\mathcal{U}$…. If $(m,q)=(2,2)$, however, the group so defined is <strong>not</strong> $\Omega_4^+(2)$, and one needs to define it in a completely different way (see Kleidman and Liebeck, p.31).</p>
<h1 id="finding-some-elements-in-omega_4q">Finding some elements in $\Omega_4^+(q)$.</h1>
<p>Let $(e_1,f_1)$ and $(e_2, f_2)$ be hyperbolic pairs, and consider the basis $\mathcal{B}={e_1, e_2, f_2, f_1}$ maintaining order. Then, one can calculate directly that $O_4^+(q)$ contains elements which are written with respect to $\mathcal{B}$ in the form</p>
\[\begin{pmatrix}
1 & a & b & ab \\ & 1 & & b \\ & & 1 & a \\ & & & 1
\end{pmatrix};\]
<p>these elements form a subgroup $S$ of size $q^2$; indeed, for $q$ odd, they form a Sylow $p$-subgroup of $O_4^+(q)$.</p>
<p>When $q$ is odd, one can look at orders to see directly that $S$ is actually a subgroup of $\Omega_4^+(q)$. Indeed, the same is true when $q$ is even, but to see this it is easiest to observe that $S$ is normalized by the element</p>
\[g=\begin{pmatrix}
1 & & & \\ & & 1 & \\ & 1 & & \\ & & & 1
\end{pmatrix}.\]
<p>This element clearly takes one element $U=\langle e_1, e_2\rangle \in \mathcal{U}_2$ to another element $W=\langle e_1, f_2\rangle\in \mathcal{U}_2$. What is more $U\cap W$ has codimension $1$ in both $U$ and $W$. This allows us to conclude that $g\in SO_4^+(q)\setminus \Omega_4^+(q)$ (see Kleidman and Liebeck, p. 30). Now order considerations imply that $S$ must be a subgroup of $\Omega_4^+(q)$.</p>
<h1 id="finding-all-elements-in-omega_4q">Finding all elements in $\Omega_4^+(q)$.</h1>
<p>Let $R$ be the set of elements whose transpose is in $S$. One sees immediately that both $R$ and $S$ lie in $\Omega_4^+(q)$ and hence so does $\langle R,S\rangle$.</p>
<p>If one sets $a$ to equal $0$, while $b$ ranges across $\mathbb{F}_q$, in both $R$ and $S$, then one immediately observes a copy of $SL_2(q)$ in $\Omega_4^+(q)$. The same is true setting $b$ to equal $0$. Since these two copies effectively “avoid interaction” one immediately obtains a copy of $SL_2(q)\circ SL_2(q)$ inside $\Omega_4^+(q)$. (The central product is due to the fact that both copies of $SL_2(q)$ share $-I$ as an element.)</p>
<p>Now one can observe that $\Omega_4^+(q)$ also contains the element</p>
\[\begin{pmatrix}
& & & 1\\ & & 1 & \\ & 1 & & \\ 1 & & &
\end{pmatrix}\]
<p>Now order considerations allow us to conclude that $\Omega_4^+(q)\cong (SL_2(q)\circ SL_2(q)):2$, and we have written down all elements of the group.</p>
<h2 id="sp_n-22a--omega_n2a">$Sp_{n-2}(2^a) < \Omega_n^+(2^a)$</h2>
<p>When $q$ is even, $\Omega_n^+(q)$ has a maximal subgroup – the stabilizer of a non-degenerate $1$-space – isomorphic to $Sp_{n-2}(q)$. I will write down the elements of this subgroup for $n=4$; the general case follows similarly.</p>
<p>First, we adjust our basis from before to be $\mathcal{C}={z_1, x_1, x_2, y_2}$ where $z_1=x_1+y_1$.</p>
<p>With respect to this basis, our quadratic form becomes</p>
\[Q(a,b,c,d) = ab+a^2+cd\]
<p>and we see that $z_1$ is non-singular. Now simply observe that the following elements stabilize $z_1$:</p>
\[\begin{pmatrix}
1 & & & c \\ & 1 & & \\ & c & 1 & c^2 \\ & & & 1
\end{pmatrix}.\]
<p>Doing the “transpose trick” like before, one immediately obtains a copy of $SL_2(q)$ in this stabilizer and (using the perfectness of $SL_2(q)$ for $q>3$ if necessary), one obtains that the stabilizer of $z_1$ in $\Omega_4^+(q)$ is isomorphic to $SL_2(q)\cong Sp_{2}(q)$.</p>
<p>One can do this more generally for larger $n$ – one simply has to exhibit the root groups of $Sp_{n-2}(q)$ inside the stabilizer of $z_1$ in $\Omega_4^+(q)$. There are two sorts of root group here: the first already ocur in $\Omega_{n-2}^+(q)$ and so are easy to write down in $\Omega_n^+(q)$; the second all take the form given above.</p>nickgillA finite classical group is best thought of as a group of linear operators on some vector space defined over a finite field. Which means, of course, that I can take a basis for this vector space and then represent the elements of my group as matrices. However, I have almost never seen anyone do this in the literature. “Taking a basis” is thought of as a rather crude thing to do when doing linear algebra – one generally tries to write general arguments that do not refer to any particular basis. However, speaking for myself, when I’m trying to work out what the hell is going on inside a finite matrix group, I often end up trying to write down individual elements as matrices… And then hiding all this when it comes to writing up the paper! This means that I have never publicly written down any of these calculations, despite using many of them over and over again. So this page is designed to be a little repository for me to note down interesting observations about such calculations… And perhaps they’ll be useful for someone else some time… Working with $\Omega_4^+(q)$ This family of groups has some weird behaviour, especially when $q$ is even. For instance, let us write $\mathcal{U}$ for the set of maximal totally singular subspaces in a formed space of type $O+$ and dimension $2m$. If $q$ is even, then, provided $(m,q)\neq (2,2)$, we can define $\Omega_{2m}^+(q)$ to be the group inducing odd permutations on $\mathcal{U}$…. If $(m,q)=(2,2)$, however, the group so defined is not $\Omega_4^+(2)$, and one needs to define it in a completely different way (see Kleidman and Liebeck, p.31). Finding some elements in $\Omega_4^+(q)$. Let $(e_1,f_1)$ and $(e_2, f_2)$ be hyperbolic pairs, and consider the basis $\mathcal{B}={e_1, e_2, f_2, f_1}$ maintaining order. Then, one can calculate directly that $O_4^+(q)$ contains elements which are written with respect to $\mathcal{B}$ in the form \[\begin{pmatrix} 1 & a & b & ab \\ & 1 & & b \\ & & 1 & a \\ & & & 1 \end{pmatrix};\] these elements form a subgroup $S$ of size $q^2$; indeed, for $q$ odd, they form a Sylow $p$-subgroup of $O_4^+(q)$. When $q$ is odd, one can look at orders to see directly that $S$ is actually a subgroup of $\Omega_4^+(q)$. Indeed, the same is true when $q$ is even, but to see this it is easiest to observe that $S$ is normalized by the element \[g=\begin{pmatrix} 1 & & & \\ & & 1 & \\ & 1 & & \\ & & & 1 \end{pmatrix}.\] This element clearly takes one element $U=\langle e_1, e_2\rangle \in \mathcal{U}_2$ to another element $W=\langle e_1, f_2\rangle\in \mathcal{U}_2$. What is more $U\cap W$ has codimension $1$ in both $U$ and $W$. This allows us to conclude that $g\in SO_4^+(q)\setminus \Omega_4^+(q)$ (see Kleidman and Liebeck, p. 30). Now order considerations imply that $S$ must be a subgroup of $\Omega_4^+(q)$. Finding all elements in $\Omega_4^+(q)$. Let $R$ be the set of elements whose transpose is in $S$. One sees immediately that both $R$ and $S$ lie in $\Omega_4^+(q)$ and hence so does $\langle R,S\rangle$. If one sets $a$ to equal $0$, while $b$ ranges across $\mathbb{F}_q$, in both $R$ and $S$, then one immediately observes a copy of $SL_2(q)$ in $\Omega_4^+(q)$. The same is true setting $b$ to equal $0$. Since these two copies effectively “avoid interaction” one immediately obtains a copy of $SL_2(q)\circ SL_2(q)$ inside $\Omega_4^+(q)$. (The central product is due to the fact that both copies of $SL_2(q)$ share $-I$ as an element.) Now one can observe that $\Omega_4^+(q)$ also contains the element \[\begin{pmatrix} & & & 1\\ & & 1 & \\ & 1 & & \\ 1 & & & \end{pmatrix}\] Now order considerations allow us to conclude that $\Omega_4^+(q)\cong (SL_2(q)\circ SL_2(q)):2$, and we have written down all elements of the group. $Sp_{n-2}(2^a) < \Omega_n^+(2^a)$ When $q$ is even, $\Omega_n^+(q)$ has a maximal subgroup – the stabilizer of a non-degenerate $1$-space – isomorphic to $Sp_{n-2}(q)$. I will write down the elements of this subgroup for $n=4$; the general case follows similarly. First, we adjust our basis from before to be $\mathcal{C}={z_1, x_1, x_2, y_2}$ where $z_1=x_1+y_1$. With respect to this basis, our quadratic form becomes \[Q(a,b,c,d) = ab+a^2+cd\] and we see that $z_1$ is non-singular. Now simply observe that the following elements stabilize $z_1$: \[\begin{pmatrix} 1 & & & c \\ & 1 & & \\ & c & 1 & c^2 \\ & & & 1 \end{pmatrix}.\] Doing the “transpose trick” like before, one immediately obtains a copy of $SL_2(q)$ in this stabilizer and (using the perfectness of $SL_2(q)$ for $q>3$ if necessary), one obtains that the stabilizer of $z_1$ in $\Omega_4^+(q)$ is isomorphic to $SL_2(q)\cong Sp_{2}(q)$. One can do this more generally for larger $n$ – one simply has to exhibit the root groups of $Sp_{n-2}(q)$ inside the stabilizer of $z_1$ in $\Omega_4^+(q)$. There are two sorts of root group here: the first already ocur in $\Omega_{n-2}^+(q)$ and so are easy to write down in $\Omega_n^+(q)$; the second all take the form given above.