(6) {\displaystyle \mathrm {G} } Quotients in the category of algebraic groups are more delicate to deal with. What is the triangle symbol with one input and two outputs? {\displaystyle n} The 1956-58 classification for algebraic groups by Chevalley over an arbitrary algebraically closed field is essentially the same as the Lie group list of Cartan-Killing. {\displaystyle \mu _{n}} To subscribe to this RSS feed, copy and paste this URL into your RSS reader. ( Let K be the connected subgroup of G such that Lie ( K) is the fixed point locus of the Cartan involution on Lie ( G ). G {\displaystyle k} k By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. n {\displaystyle \mathrm {G} /\mathrm {H} } {\displaystyle k} Not all algebraic groups are linear groups or abelian varieties, for instance some group schemes occurring naturally in arithmetic geometry are neither. Namely, from our discussion above the difference is precisely measured by $H^1(\mathbb{R},G)$. Asking for help, clarification, or responding to other answers. .[4]. [ The answer to this is yes, although the proof in complete generality is more difficult than one might first imagine. [5] Note that if the field The simply connected groups you list are simple algebraic groups and are simple as an abstract group under the conditions I gave in my answer. and, in fact, $Z(G)(\mathbb{R})=Z(G(\mathbb{R}))$ (e.g. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. {\displaystyle k} DOI: 10.1016/j.jalgebra.2020.06.006 Corpus ID: 215814287; Adjoint algebraic groups as automorphism groups of a projector on a central simple algebra @article{Petrov2020AdjointAG, title={Adjoint algebraic groups as automorphism groups of a projector on a central simple algebra}, author={Viktor Petrov and Andrei Yu. ( Asking for help, clarification, or responding to other answers. x Namely, let us denote by $K$ the image of $K$ under the map $G(\mathbb{R})\to G^\mathrm{ad}(\mathbb{R})$. So, let $G$ be a reductive group over $\mathbb{R}$. The adjoint group $\Ad G$ is contained in the group $\def\Aut{\mathop{\textrm{Aut}}} \def\g{\mathfrak g} \Aut \g $ of automorphisms of the Lie algebra $\g$ of $G$, and its Lie algebra coincides with the adjoint algebra $\Ad\g$ of $\g$. such that However these two groups are not isomorphic as algebraic group (only isogenous). Jay, thank you very much for your answer. ( 1 \rightarrow Z_G \rightarrow G \rightarrow G^{\operatorname{ad}} \rightarrow 1 Jim. {\displaystyle \mathrm {GL} _{n}} is a G . $$ , its image is an algebraic subgroup of So, from this, we see that your question becomes: is $Z(G^\mathrm{ad})=0$? and Abstract-algebra . Stack Overflow for Teams is moving to its own domain! Use MathJax to format equations. {\displaystyle \mathrm {G} } We will construct the wonderful compacti cation of a semisimple algebraic group of adjoint type G, following mostly the well-known survey of Evens and Jones [EJ]. @Demin: for each given type there is a simply connected algebraic group of this type, unique up to isomorphism. The latter are classified over algebraically closed fields via their Lie algebra. Thanks for contributing an answer to Mathematics Stack Exchange! the linear group $\def\Ad{\mathop{\textrm{Ad}}} \Ad G$ which is the image of the Lie group or algebraic group $G$ under the adjoint representation (cf. G {\displaystyle \mathrm {G} \times \mathrm {G} \to \mathrm {G} } If the field Note that this composite map doesnot have any kernel over $\mathbb{C}$. {\displaystyle \pi :\mathrm {G} \to \mathrm {G} /\mathrm {H} } Also, see Chevalley's Collected Works here. asked Dec 8, 2013 at 8:54. Slick Hybrid Bike Tires on Steep Gravel Descent? k In mathematics, an algebraic group is an algebraic variety endowed with a group structure which is compatible with its structure as an algebraic variety. How to grow a Dracaena from a broken branch, Snubber capacitor vs RC snubber (power supply), English Tanakh with as much commentary as possible. In order to use results from semi-simple Lie groups for the reductive group G ( R), it seems common to implicitly use the adjoint group of G. If I understand correclty, it is defined by the exact sequence of algebraic groups. For $\mathsf E_6$ and $\mathsf E_7$, the adjoint form (arising as the automorphism group of the Lie algebra) has fundamental group that is cyclic of prime order (3 and 2, respectively). In order to use results from semi-simple Lie groups for the reductive group $G(\mathbb{R})$, it seems common to implicitly use the adjoint group of $G$. {\displaystyle k} For an example: in characteristic 2 the symplectic group $Sp_{2n}$ is simple as an abstract group as it is isomorphic (as an abstract group) to the corresponding adjoint group. For example the additive group can be embedded in k Press (1958) (Translated from Russian), J.-P. Serre, "Lie algebras and Lie groups", Benjamin (1965) (Translated from French). This is a group topology, and it makes Adjoint representation of a Lie group). This page was last edited on 20 January 2022, at 17:24. Reductive group schemes. see [1, Theorem 17.93]). ( To learn more, see our tips on writing great answers. carries a Zariski topology. n , respectively, into If Linearity of maximum function in expectation. G H @Demin: The question is "simple" (so to speak) but takes extra thought in prime characteristic. {\displaystyle \mathrm {H} } see [2, Theorem D.2.8]) and thus so is $\widetilde{K}$. rev2022.11.14.43032. k Abelian varieties are connected projective algebraic groups, for instance elliptic curves. G Share. and a surjective morphism {\displaystyle x\mapsto \left({\begin{smallmatrix}1&x\\0&1\end{smallmatrix}}\right)} When the migration is complete, you will access your Teams at stackoverflowteams.com, and they will no longer appear in the left sidebar on stackoverflow.com. It turns out that the classification of semi-simple algebraic groups over an arbitrary algebraically closed field $K$ is analogous to the case $K=\C$, that is, a semi-simple algebraic group is determined up to . An algebraic group is said to be connected if the underlying algebraic variety is connected for the Zariski topology. Levi's theorem states that every such is (essentially) a semidirect product of a unipotent group (its unipotent radical) with a reductive group. 1 ! G But, in general, they wont be equal. This happens with $G=\mathrm{SL}_2$. If this answers your question it would be good to accept the answer so people know the question is answered with. k The European Mathematical Society, 2010 Mathematics Subject Classification: Primary: 20GXX Secondary: 14LXX [MSN][ZBL]. What are some examples of non-reductive groups? In particular, we see that, $$\mathrm{Ad}_\mathbb{R}(G(\mathbb{R}))\cong G(\mathbb{R})/Z(G)(\mathbb{R})$$. By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. ( Let G be a (not necessarily quasi-split) reductive group over Q. Indeed, note that $\widetilde{K}^+$ is finite index in $\widetilde{K}$ and $K^+$ is finite index in $K$ and the map $G(\mathbb{R})^+\to G^\mathrm{ad}(\mathbb{R})^+$ is surjective (by prior discussion) from where the claim follows. You will sometimes see such naming conventions used, especially for finite reductive groups. We also compute the essential dimension of orthogonal and . According to Yves comment, in each type of your list there is exactly one simply connected group. Structure theorem for general algebraic groups, Algebraic groups over local fields and Lie groups, Affine Group Schemes; Lie Algebras; Lie Groups; Reductive Groups; Arithmetic Subgroups, https://en.wikipedia.org/w/index.php?title=Algebraic_group&oldid=1115202038, Short description is different from Wikidata, Pages that use a deprecated format of the math tags, Creative Commons Attribution-ShareAlike License 3.0. orthogonal and symplectic groups are affine algebraic groups. {\displaystyle k} This is particularly nice since $H^1(\mathbb{R},Z(G))$ is a finite abelian group which can be explicitly calculated (e.g. Question 3: Lets try to split our understanding into two parts. G They are always commutative. It is also going to be finite index. Deep relations between linear algebraic groups over an arbitrary field and central simple algebras with involution can be traced back to two main sources. The simply connected groups are those groups for which the weight lattice of the root system of $G$ is equal to $X$; this is the same as those groups of each type with the `largest' (finite) centre. In general assume G is connected reductive algebraic group and let : G G a d be an adjoint quotient of G. Indeed, note that if $G(\mathbb{R})$ is connected, then so is $K$ (e.g. $$, $G(\mathbb{R}) / C_{G(\mathbb{R})}\,(G(\mathbb{R})^\circ)$, $\tilde{K} \subset G^{\operatorname{ad}}(\mathbb{R})$, $Z(O(2)) = Z_{GL_2} \, ((\mathbb{R})) \cap O(2) = \{ \pm 1 \}$, $H^1(\mathbb{R},G):=H^1(\mathrm{Gal}(\mathbb{C}/\mathbb{R}),G(\mathbb{C})$, $G^\mathrm{ad}(\mathbb{R})=G(\mathbb{R})/Z(G)(\mathbb{R})$, $\mathrm{Ad}:G\to \mathrm{GL}(\mathrm{Lie}(G))$, $G(\mathbb{R})\to G^\mathrm{ad}(\mathbb{R})$, $G(\mathbb{R})^+\to G^\mathrm{ad}(\mathbb{R})^+$, $Z(G)(\mathbb{R})\to G^\text{der}(\mathbb{R})\to G(\mathbb{R})$. ! P Then $G$ is simple as an abstract group if the kernel of $\pi$ is trivial because $\pi$ is then an isomorphism of abstract groups. {\displaystyle \mathrm {G} } is isomorphic to its adjoint group) if and only if its roots generate the lattice of rational characters of the maximal torus; the centre of such a group is trivial. (that is, the maps {\displaystyle \mathrm {H} } In addition it is both affine and projective. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. In mathematics, an algebraic group is an algebraic variety endowed with a group structure which is compatible with its structure as an algebraic variety. k G , and the number of elements of the general linear group over a finite field is the q-factorial You might find section 1.11 in Carter's book "Finite Groups of Lie Type" a useful read as he reviews the classification of simple algebraic groups. Let G be a simple algebraic group of adjoint type over an algebraically closed field K of characteristic p > 0. Connect and share knowledge within a single location that is structured and easy to search. k Why is there "n" at end of plural of meter but not of "kilometer". First observe that if p divides the order of x then C G ( x ) is G . Algebraic groups: The theory of group schemes of finite type over a field (Vol. Thus a simple algebraic group Gwhose Lie algebra is isomorphic to g and whose rational simple modules (also denoted L( )) all have such highest weights is precisely a group of adjoint type. A toy example would maybe be $GL_2$ with maximal compact subgroup $O(2) \subset GL_2(\mathbb{R})$. EDIT: I wrote the below quickly, so I would double-check it for correctness. In my opinion it would be reasonable to give them a special name, for example $E_6^{sc}$. The adjoint group $\Ad G$ is contained in the group $\def\Aut{\mathop{\textrm{Aut}}} \def\g{\mathfrak g} \Aut \g $ of automorphisms of the Lie algebra $\g$ of $G$, and its Lie algebra coincides with the adjoint algebra $\Ad\g$ of $\g$. How can creatures fight in cramped spaces like on a boat? reference-request. We give a lower bound for the essential dimension of a split simple algebraic group of "adjoint" type over a field of characteristic 2. Thus, in particular for classification purposes, it is natural to restrict statements to connected algebraic group. 1 and This connectivity hypothesis is, shockingly, satisfied if $G$ is simply connected as an algebraic group (i.e. G For simplicitly, Im going to assume that $G$ is semisimple. a finite adjoint Chevalley group or twisted type) and / is an irreducible discrete series character of G, then the restriction of / to U (a maximal unipotent subgroup . By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. is a subvariety But I'm still confused about the terminology: In contrast to finite groups, where each simple group has a definite name, simple algebraic groups doesn't seem to have a name on their own (apart from the classical types). More generally every finite group is an algebraic group (it can be realised as a finite, hence Zariski-closed, subgroup of some The results obtained in the study of operator algebras are phrased in algebraic terms, while the techniques used are . k In your list, SO must be replaced by Spin. Great answer, thank you very much! In functional analysis, a branch of mathematics, an operator algebra is an algebra of continuous linear operators on a topological vector space, with the multiplication given by the composition of mappings . (1) Is this group also simple ? (2) Do you happen to know the simply connected groups of the listed types ? As a vector space the Lie algebra is isomorphic to the tangent space at the identity element. Do solar panels act as an electrical load on the sun? G {\displaystyle \mathbb {C} } H defining the group structure map One definition is that a connected linear algebraic group G over a perfect field is reductive if it has a representation with finite kernel which is a direct sum of irreducible representations. Why would an Airbnb host ask me to cancel my request to book their Airbnb, instead of declining that request themselves? / by the morphism If you have a burning desire to know what happens in the general case, I can return later. Two main re-sults are proved: the first is that under appropriate conditions, all nonlinear irre- . Thanks for contributing an answer to Mathematics Stack Exchange! In type $B_n$, is $s_\beta(\alpha)=\alpha+\sqrt{2}\beta$? (the neutral element), and regular maps Cambridge University Press. th roots of unity in the multiplicative group If $G$ is an adjoint algebraic group then it is always simple as an abstract group, (EDIT: This is because any proper normal subgroup of a simple algebraic group must be finite and lie in the centre). {\displaystyle \mathrm {G} } Then, by the above we have that, $$(G(\mathbb{R})/Z(G)(\mathbb{R}))^+\to G^\mathrm{ad}(\mathbb{R})^+$$. For nontrivial representations the kernel is finite, hence central (by normality). Indeed, if $H$ has non-trivial center then $H\to H/Z(H)$ is a non-trivial isogeny (note $Z(H)$ is finite since $H$ is semisimple). G ( is that it is that of a group scheme over results from algebraic group theory over algebraically closed fields. An algebraic group $G$ is semi-simple if and only if $G$ is a product of simple connected closed normal subgroups. Making statements based on opinion; back them up with references or personal experience. This implies for example that $G^{\operatorname{ad}}$ is a direct product of its simple factors). = Let G be a simple exceptional algebraic group of adjoint type over an I have corrected my answer accordingly. G Lie group, semi-simple). $G_2$, $F_4$ or $E_8$ any characteristic. / What happends with the ownership of land where the land owner no longer exsists? So there is just one simply connected group of each simple type. G (e.g. Many matrix groups are also algebraic. This article was adapted from an original article by A.L. G where $\sigma$ is the non-trivial element of $\mathrm{Gal}(\mathbb{C}/\mathbb{R})$ acting on $G(\mathbb{C})/Z(G)(\mathbb{C})$ so in particular, an element of $G^\mathrm{ad}(\mathbb{R})$ neednt come from an element of $G(\mathbb{R})$ since these are elements of $G(\mathbb{C})$ which are fixed by $\sigma$, whereas elements of $G^\mathrm{ad}(\mathbb{R})$ are represented by elements of $G(\mathbb{C})$ which neednt be fixed by $\sigma$ on the nose but, instead, fixed up to multiplication by an element of $Z(G)(\mathbb{C})$. k Then x and C G ( x ) are as in Table ; we label x according to its order,which is not divisible by p .Proof. But, if $G^\mathrm{ad}(\mathbb{R})$ is disconnected then so is $\widetilde{K}$ (again by [2, Theorem D.2.8]) and so youre in big trouble. So, you can deduce from this that, roughly, the same argument in the semisimple case applies $K$ and $\widetilde{K}$ differ by finite index. Way to create these kind of "gravitional waves". G closed field with no assumption on the characteristic) and of which type are they ? Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Yet another definition of the concept is to say that an algebraic group over If not, please let me know and I will delete the question that emerged when I was trying to understand some theorems in a paper on the vanishing range of the cohomology of finite groups of Lie type. An algebraic group is said to be affine if its underlying algebraic variety is an affine variety. Let $G$ be a (not necessarily quasi-split) reductive group over $\mathbb{Q}$. 15. It is not in general a group topology, i.e. Note that the image of $K\cap G^\mathrm{der}(\mathbb{R})$ in $G^\mathrm{ad}(\mathbb{R})$ is contained in $K$ up to finite index, because the product map $Z(G)(\mathbb{R})\to G^\text{der}(\mathbb{R})\to G(\mathbb{R})$ has finite cokernel and kernel (with cokernel equal to $H^1(\mathbb{R},Z(G^\text{der}))$). that $G(\mathbb{C})$ is a simply connected complex Lie group)this follows form [1, Theorem 25.54]). Preamble: In particular, if $G$ is semi-simple, $\Ad G$ coincides with the connected component of the identity in $\Aut \g$. 1.3 We should emphasize that in the literature cited here, there is some variation in the initial set-up: sometimes one has Lie (or algebraic) groups rather . semisimple algebraic group of adjoint type dened over a perfect eld k. Suppose the longest element w 0 of the Weyl group W(G,T) acts by 1 on the set of roots with respect to a xed maximal torus T. Then for a strongly regular element t G(k), we prove that t is real in G(k) if and only if t is strongly real in G(k) (Theorem 2.1.2). Let G be a semisimple algebraic group of adjoint type over the field of complex numbers. {\displaystyle \mathrm {G} (k)\to \mathrm {G} (k)/\mathrm {H} (k)} {\displaystyle \mathrm {G} \to \mathrm {G} '} {\displaystyle \mathrm {H} } Thanks for contributing an answer to MathOverflow! Many groups of geometric transformations are algebraic groups; for example, orthogonal groups, general linear groups, projective groups, Euclidean groups, etc. Similarly to the Lie groupLie algebra correspondence, to an algebraic group over a field Before asking the question I should say that I don't know much about algebraic groups and I'm not sure if the question has the right level for MO. see [2, Theorem D.2.8]). If ( W, S) is the Coxeter system and s S, and s is a simple root with corresponding coroot s , is it true that s ( 1) = 1? If $G$ is absolutely simple simply connected, why is G(F_v) quasisimple for almost every valuation v? Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company. [10] Chevalley's structure theorem asserts that every connected algebraic group is an extension of an abelian variety by a linear algebraic group. ) {\displaystyle \mathrm {G} } Since $\operatorname{Ad}$ is continuous, the image $\operatorname{Ad}(K)$ should be compact in $\operatorname{Ad}(G(\mathbb{R}))$, but I see no reason for it to be maximal. Making statements based on opinion; back them up with references or personal experience. G is the kernel of Confused by reflection formula for Coxeter systems of type $I_2(4)$ and $I_2(6)$ in Humphreys. @Demin As Jim says the simply connected groups of types $B_n$ and $D_n$ are $Spin_{2n+1}$ and $Spin_{2n}$ respectively (they are double covers of the corresponding special orthogonal groups). It only takes a minute to sign up. A connected semi-simple group is a group of adjoint type (i.e. is an isomorphism onto $G^\mathrm{ad}(\mathbb{R})^+$. Look at the co-charactre in $SL_2$ and that doesnot satisfy what you have written. Chain Puzzle: Video Games #02 - Fish Is You, The meaning of "lest you step in a thousand puddles with fresh socks on". n As an algebraic variety {\displaystyle [n]_{q}!} H G {\displaystyle k} English Tanakh with as much commentary as possible. x When proving certain theorems about connected reductive algebraic groups you reduce to the case where G is simple and either adjoint or simply connected. Asking for help, clarification, or responding to other answers. {\displaystyle k} Can anyone give me a rationale for working in academia in developing countries? Pontryagin, "Topological groups", Princeton Univ. Why are open-source PDF APIs so hard to come by? is not algebraically closed, the morphism of groups Since people tend to think about the notion of simply connected in topological terms for Lie groups, it's always a good idea to add a reminder about the meaning in the algebraic setting of Chevalley's seminar on classification. ; thus the symmetric group behaves as though it were a linear group over "the field with one element". The adjoint group of a linear group $G$ is the linear group $\def\Ad{\mathop{\textrm{Ad}}} \Ad G$ which is the image of the Lie group or algebraic group $G$ under the adjoint representation (cf. $$G^\mathrm{ad}(\mathbb{R})=(G(\mathbb{C})/Z(G)(\mathbb{C}))^\sigma$$. G Thus you can find a co-character such that it doesnot satisfy your condition. I believe for the classical types these are $\text{SL}_{n+1}$ $(A_n)$, $\text{SO}_{2n+1}$ $(B_n)$, $\text{Sp}_{2n}$ $(C_n)$, $\text{SO}_{2n}$ $(D_n)$. {\displaystyle k} What is the mathematical condition for the statement: "gravitationally bound"? ) Some nice, introductory notes are available here. ) {\displaystyle k} n rev2022.11.14.43032. Operator algebra. Finally, to understand the difference between $K$ and $K$ we note that since $Z(G)(\mathbb{R})$ is compact its contained in every $K$ and so $K=K/Z(G)(\mathbb{R})$. George, thanks for your good explanation and the references. Let be a simple simply-laced algebraic group of adjoint type over the field of complex numbers, be a Borel subgroup of containing a maximal torus of In this article, we show that is a minuscule fundamental weight if and only if for any parabolic subgroup containing properly, there is no Schubert variety in such that the minimal parabolic subgroup of is the connected component, containing the identity automorphism of the group of all algebraic automorphisms of. L may not be surjective (the default of surjectivity is measured by Galois cohomology). G Using the action of an affine algebraic group on its coordinate ring it can be shown that every affine algebraic group is a linear (or matrix group), meaning that it is isomorphic to an algebraic subgroup of the general linear group. 1 \rightarrow Z_G \rightarrow G \rightarrow G^{\operatorname{ad}} \rightarrow 1 Now one can embed $SL_2 \hookrightarrow SL_3$ as the first $2 \times 2$ block followed by the map to the adjoint group of $SL_3$. It only takes a minute to sign up. (the inversion operation) which satisfy the group axioms.[2]. If G is an adjoint algebraic group then it is always simple as an abstract group, (EDIT: This is because any proper normal subgroup of a simple algebraic group must be finite and lie in the centre). {\displaystyle n!} @Demin: You need to do more background reading in Lie theory. {\displaystyle \mathrm {G} \to \mathrm {G} } G is isomorphic to its adjoint group) if and only if its roots generate the lattice of rational characters of the maximal torus; the centre of such a group is trivial. If the ground field is the field $\C$ of complex numbers, a semi-simple algebraic group is nothing but a semi-simple Lie group over $\C$ (cf. {\displaystyle \mathrm {G} ,\mathrm {G} '} {\displaystyle \mu _{2}} A 1-TYPE SUBGROUPS CONTAINING REGULAR UNIPOTENT ELEMENTS TIMOTHY C. BURNESS AND DONNA M. TESTERMAN Abstract. Conversely, if $f:H\to H$ is a non-trivial isogeny then $\ker(f)$ is a non-trivial subgroup of $H$ and, in fact, its contained in $Z(H)$. The reason is that $\ker(f)$ is finite and so the conjugation action $H\to \mathrm{Aut}(\ker(f))$ has trivial image since $\mathrm{Aut}(\ker(f))$ is a finite group (scheme) and $H$ is connected. {\displaystyle \mathrm {G} (k)} Autour des schmas en groupes, 1, pp.93-444. is a normal algebraic subgroup of {\displaystyle n\geq 1} (group schemes can more generally be defined over commutative rings). Semantic Scholar extracted view of "Linear Algebraic Groups and Finite Groups of Lie Type" by Gunter Malle et al. ) MathJax reference. {\displaystyle \mathrm {G} (k)} Added: The mathematical problem is solved by Jay's answer. For instance people sometimes write $E_6^{sc}(q)$. [2] Conrad, B., 2014. -group then the group {\displaystyle k=\mathbb {R} } I would imagine the reason for the way it's written is the following. Onishchik (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. https://encyclopediaofmath.org/index.php?title=Adjoint_group&oldid=51930, N. Bourbaki, "Elements of mathematics. Thus the study of algebraic groups belongs both to algebraic geometry and group theory. Suppose $G$ is a semisimple algebraic group of adjoint type. [9] If can be made very explicit in some cases, for example over the real or p-adic fields, and thereby over number fields via local-global principles. k as a quasi-projective variety. Thank you! the group operations may not be continuous for this topology (because Zariski topology on the product is not the product of Zariski topologies on the factors[11]). ( H H (each point is a Zariski-closed subset so it is not connected for I would be grateful for any answer. Let G be a simple algebraic R -matrix group of adjoint type and suppose that there exists a morphism h : S ( R) G satisfying the two conditions M and S1. Let X = PSL 2 (q) be a subgroup of G, where q 4 is a p-power, and let x X be an . {\displaystyle \mathbb {P} ^{n}(k)} Question 1: Ok, so let $\mathrm{Ad}:G\to \mathrm{GL}(\mathrm{Lie}(G))$ be the adjoint representation. Abstract. Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. k Many matrix groups are also algebraic.

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