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Let $G$ be a (connected) reductive group over some ground field $F$ and $G^*$ its unique quasi-split inner form. Denote by $\operatorname{rank}_F G$ the split rank of $G$, i.e. the dimension of a maximal $F$-split torus in $G$, and likewise for $G^*$. Is it true that

$$\operatorname{rank}_F G\le \operatorname{rank}_FG^*$$

with equality holds only if $G$ is isomorphic to $G^*$?

For example, this will say that among unitary group that splits over a fixed separable quadratic extension $E/F$, the quasi-split one has the largest split rank (which is $\lfloor\frac{n}{2}\rfloor$ for $U_n$), and is the only unitary group that achieves this split rank. For split groups this result will be obvious.

I am mostly wondering about the case when $F$ is a non-archimedean local field (for Local Langlands, where the same Langlands parameter can be attached representations of both $G$ and $G^*$), but any suggestion/reference will be great. Thanks!

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    $\begingroup$ To make this question a little more concrete, maybe a famliar example such as an $F$-split simple group $G$ of type $A_n$ would be helpful? (Anyway I guess you are following the language of Borel-Tits here, as in the Tits classification.) $\endgroup$ – Jim Humphreys Jan 26 '18 at 0:25
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    $\begingroup$ I am worried that the variety of languages I know is limited (i.e. this is the only language I am familiar with). But thanks and let me try to add some elaborations and example to the question. $\endgroup$ – Cheng-Chiang Tsai Jan 26 '18 at 1:31
  • $\begingroup$ To go through the classical groups might give an interesting data-point set, since, obviously, no general fact that fails (for relatively elementary reasons, as opposed to general algebraic-group reasons) there can be true generally. It would engender some intuition, I'd think, though, who knows, the exceptional groups can always do weird things. Even D4 and "triality". Still, classical groups are a good test sample... $\endgroup$ – paul garrett Jan 26 '18 at 1:37
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    $\begingroup$ At least one can check that this is true for absolutely simple groups, by just going through and comparing the table of indices in Tits article "Classification of algebraic semisimple groups" and in Springer book "Linear Algebraic groups" (by noting that when Springer says the index appears for outer type, than it means for you it's an inner form of a non-split quasi-split algebraic group). This leaves out two things to do: give a conceptual explanation of this fact, and then understand how tori interfere in this situation. $\endgroup$ – thierry stulemeijer Jan 26 '18 at 1:56
  • $\begingroup$ @JimHumphreys, since the question is about split ranks, surely considering an already $F$-split group $G$, where $G = G^*$, does not shed any additional light? $\endgroup$ – LSpice Jan 26 '18 at 3:12
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Every torus in $G$ transfers to $G^*$, so we definitely have the desired inequality. If we have equality, then there is a maximal split torus $A$ in $G$ that is also maximal split when transferred to the torus $A^*$ in $G^*$; so $C_{G^*}(A^*)$ is a torus; so $C_G(A)$, which is isomorphic over the separable closure to $C_{G^*}(A^*)$, is a torus; so $G$ is quasisplit, hence isomorphic to $G^*$.

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    $\begingroup$ Loren, I believe you have exactly pointed out what I don't know! How do we have that every torus in $G$ transfers to $G^*$? $\endgroup$ – Cheng-Chiang Tsai Jan 26 '18 at 3:20
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    $\begingroup$ This is a beautiful fact that I've been waiting to use ever since Stephen pointed it out to me this summer. It is due to Raghunathan; see mathscinet.ams.org/mathscinet-getitem?mr=2125504 . $\endgroup$ – LSpice Jan 26 '18 at 3:29
  • $\begingroup$ The ref: Raghunathan, M. S. Tori in quasi-split-groups. J. Ramanujan Math. Soc. 19 (2004), no. 4, 281-287. $\endgroup$ – YCor Jan 26 '18 at 3:43
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    $\begingroup$ The result is actually due to Langlands. See Lemma 2.1 of his 'On the classification of representations of real algebraic groups'. It's also probably worth noting that the main content is a theorem of Steinberg on the existence of rational points in rational conjugacy classes of simply connected semisimple groups. $\endgroup$ – Keerthi Madapusi Pera Jan 28 '18 at 4:38
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    $\begingroup$ @LSpice As I mentioned, the only non-trivial input is Steinberg's theorem on the existence of rational representatives of conjugacy classes in simply connected, quasi-split groups. This is Theorem 1.7 of his paper 'Regular elements of semisimple algebraic groups' and holds for any perfect field. So Langlands' proof definitely applies to all char 0 non-arch local fields. I'm not sure what happens in char p. $\endgroup$ – Keerthi Madapusi Pera Jul 9 '18 at 8:54
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Whilst trying to understand the answer of Loren, I finally came up with an elementary explanation (no need for Steinberg’s result). Since it also gives another picture of the situation, I think it’s worth writing.

Let $G$ and $G^*$ be algebraic groups over $F$ as in the question. Let $K$ be a finite Galois extension of $F$ splitting those groups, and let $G_0\cong G_K\cong (G^*)_K$ be the corresponding split group over $K$. Let $\varphi \colon G_0\to (G^*)_{K}$ and $\psi \colon G_0\to (G)_{K}$ be the given isomorphisms. With those notations, $G^*$ (respectively $G$) is obtained from $G_0$ by twisting against the cocycle $A_{\sigma} = \varphi^{-1}(^{\sigma}\varphi ) $ (respectively $B_{\sigma} = \psi^{-1}(^{\sigma}\psi )$), where $\sigma \in \Gamma = $Gal$(K/F)$.

We can prove that $Z(G^*)$ and $Z(G)$ are isomorphic over $F$. Indeed, they are both $K/F$ forms of $Z(G_0)$, where the twisting has values in Aut $G_0 \cong G_0/Z(G_0)\rtimes $Out $G_0$. Hence, $Z(G^*)$ and $Z(G)$ only depends on the image of $A_{\sigma}$ and $B_{\sigma}$ under $H^1(\Gamma ,\text{Aut }G_0))\to H^1(\Gamma ,\text{Out }G_0) $. But those images are cohomologous because $G$ and $G^*$ are inner forms.

Going further, let $S$ (resp. $S^*$) be a maximal $F$-split torus in $G$ (resp. $G^*$), and let $T$ (resp. $T^*$) be a maximal $F$-torus containing it. The next claim is that $ \text{rk}_F T/Z(G)\leq \text{rk}_F T^*/Z(G^*)$. Indeed, since $G/Z(G)$ and $G^*/Z(G^*)$ are inner forms of each others, the so called $*$-action on their Dynkin diagram is the same. Hence $T/Z(G)$ cannot have more characters defined over $F$ than $T^*/Z(G^*)$. Also note that $\text{rk}_F T/Z(G) = \text{rk}_F T^*/Z(G^*)$ iff $G/Z(G)$ is quasi-split, which in turn is equivalent to $G/Z(G)\cong G^*/Z(G^*)$ by uniqueness of quasi-split forms in an inner class.


EDIT: To prove the two last sentences of the previous paragraph, we can assume that the groups are adjoint simple. Also, since Weil restriction is encoded in the $*$-action, we have $G=R_{F'/F}(G')$ and $G^* = R_{F'/F}({G^*}')$ for some unique absolutely simple $G'$ and ${G^*}'$ that are also inner forms of each other, so we can even assume that the groups are adjoint absolutely simple.

Hence, we are looking at a single Dynkin diagram with a $*$-action. Now conlcude by observing that the rank of $G^*$ is the number of orbits of the Dynkin diagram, and that the rank of $G$ is the number of orbits such that the corresponding invariant character of $T$ restricts to a non-trivial character of $S$. In terms of Tits index, this is just saying that for a quasi-split groups, all orbits are circled, while for an inner form of it, you add some anisotropicity, so that some orbits are not circled anymore. For the relation with quasi-splitness, recall that the group is quasi-split iff all orbits of the $*$-action are circled in the index.


Finally, the inequality $\text{rk}_F G\leq \text{rk}_F G^*$ holds because given an exact sequence of diagonalizable $F$-groups $1\to A\to B\to C\to 1$, $\text{rk}_F(B) = \text{rk}_F(A) + \text{rk}_F(C)$. In case of equality, as noted before, $G/Z(G)$ is quasi-split, hence so is $G^{der}$. Thus $G^{der}\cong (G^*)^{der}$, so that $G\cong G^*$, because a connected reductive group $H$ is reconstructed from the triple $(H^{der},Z(H),Z(H^{der})\hookrightarrow Z(H))$.

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  • $\begingroup$ In the second last paragraph, how do we get that rank in the adjoint group are the same iff it is the (unique) quasi-split inner form? $\endgroup$ – Cheng-Chiang Tsai Jan 31 '18 at 1:48
  • $\begingroup$ I guess that we can have a small industry of trying to understand one another's answers. :-) I don't understand "Hence $T/Z(G)$ cannot have more characters defined over $F$ than $T^*/Z(G^*)$." It seems to me that so far the discussion has been symmetric in $G$ and $G^*$ (not yet using quasi-splitness of $G^*$), and yet there is some asymmetry in this statement; what broke the symmetry (or, to put it differently, how does the statement I quoted use the quasi-splitness of $G^*$)? $\endgroup$ – LSpice Jan 31 '18 at 2:36
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    $\begingroup$ @Cheng-ChiangTsai and Loren (unfortunately, I can only ping one person per comment), thank you for your interest! I've added some explanations, I hope this makes the picture clearer. $\endgroup$ – thierry stulemeijer Jan 31 '18 at 12:37
  • $\begingroup$ Thanks! Now I see that there are some lack of knowledge of mine regarding $\ast$-action; I'll go read about it. $\endgroup$ – Cheng-Chiang Tsai Jan 31 '18 at 20:55

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