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$G$ is a connected, reductive group over a field $k$, $S$ is a maximal $k$-split torus of $G$, and $_k \Phi = \Phi(S,G)$ is the set of roots of $S$ in $G$. Equivalently, $_k \Phi$ is the restriction to $S$ of the set $\Phi = \Phi(T,G)$ of roots of $T$ in $G$, removing any roots which are trivial on $S$ (e.g. see here mathoverflow.net/questions/255283/two-definitions-of-restricted-roots).

This part in Borel, Linear Algebraic Groups, explains the relationship between certain bases of $\Phi$ and $_k \Phi$.

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I've been staring at (3) for awhile.

It is clear to me that $_k\Delta \subseteq j(\Delta)$: by construction, every element of $_k\Phi^+$ is an nonnegative integral combination of elements of $j(\Delta)$, so $j(\Delta)$ must contain the unique base of $_k \Phi^+$, namely $_k \Delta$.

But I don't understand why $j(\Delta)$ should be contained in $_k \Delta \cup \{0\}$.

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    $\begingroup$ Directly proving that $j(\Delta)$ is a linearly independent set seems to be a bit of a conundrum. If nobody comes along with a short direct proof in the context of the above reference, you can opt to take a different approach to setting up the relative root system: see sections 11-12 (esp. Warning 11.3.5 and Prop. 12.1.3, whose proof rests on the discussion in 12.2) in ams.org/open-math-notes/omn-view-listing?listingId=110663 $\endgroup$
    – nfdc23
    Apr 12, 2017 at 3:25
  • $\begingroup$ Thank you, I will try to read it. I don't think I have ever found a statement in Borel's book which was not properly justified (if we don't count the algebraic geometry review in the beginning of the book), so I am hopeful that my difficulty here is me being imperceptive $\endgroup$
    – D_S
    Apr 12, 2017 at 3:30
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    $\begingroup$ His book is great, but here are two cases of incorrect proofs there: (i) in the proof of 4.10, since $W$ is merely quasi-affine its coordinate ring $K[W]$ in the sense of $K$-algebra of global functions need not be finitely generated; (ii) the proof of the integrality property for the relative root system in 21.6 (taken from the Borel-Tits paper on reductive groups) doesn't work if $\alpha$ is a non-trivial multiple of a relative root (as can happen). Issue (i) is fixed in Prop. E.2.1 of the above link, and (ii) is bypassed in the development of relative root systems there. $\endgroup$
    – nfdc23
    Apr 12, 2017 at 4:05
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    $\begingroup$ As a special case of the need to prove $j(\Delta)$ is linearly independent (really I mean that $j(\Delta) - \{0\}$ is linearly independent), I don't see how Borel is ruling out that some element of $j(\Delta)$ is twice another (in case the relative root system is non-reduced, as can happen). This is a manifestation of error (ii) in my previous comment. $\endgroup$
    – nfdc23
    Apr 12, 2017 at 4:53
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    $\begingroup$ That result in Borel-Tits (same as what you are asking about from Borel's textbook) is correct; it is just the proof in Borel-Tits that is incorrect, but a proof is given in the link I provided (and the answer below addresses as well, though to be honest I had thought that the proof of 15.5.3(iii) in Springer's book overlooked proving the linear independence issue). I don't think Springer's discussion of the relative root system on pp. 259-260 of his book explains why the relative coroot lies in the expected dual lattice, but maybe someone can find an argument lurking in the shadows there... $\endgroup$
    – nfdc23
    Apr 13, 2017 at 2:18

1 Answer 1

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That's really strange since Borel and Tits have a detailed and non-obvious proof of $j(\Delta)\subseteq{}_k\Delta\cup\{0\}$ (Proposition 6.8) in their paper. Here is a simplification which I extracted from the corresponding statement (Prop. 15.5.3) in Springer's book.

There are two $\Gamma=Gal(K/k)$-actions on $X^*(T)$: the natural one $\gamma(\chi)$ and the $*$-action $w*\chi=w_\gamma(\gamma(\chi))$ where $w_\gamma$ is the unique Weyl group element with $\gamma(B)=w_\gamma^{-1}Bw_\gamma$. The key observation is that $w_\gamma$ is an element of the Weyl group of the anisotropic kernel $M=C_G(S)$. Let $\Delta^0\subseteq\Delta$ be the set of simple roots of $M$. Its elements are the simple roots restricting to $0$ on $S$. It follows that $$ (1)\qquad\gamma(\chi)-\gamma*\chi\in\langle\Delta^0\rangle_{\mathbb Q}. $$ Let $Y(T):=X_*(T)\otimes\mathbb Q$. Then $Y(S)=Y(T)^\Gamma$ with respect to the natural $\Gamma$-action. Using $(1)$ one sees that $Y(S)$ is defined by the equations $$ (2)\qquad\{\sigma=0\mid\sigma\in\Delta^0\}\cup\{\chi-\eta=0\mid\chi\in X^*(T),\eta\in\Gamma*\chi\}. $$ So these span the kernel of $j$. Let $\Delta'\subseteq\Delta\setminus\Delta^0$ be a set of representatives of the $\Gamma*$-orbits. Then $\Delta'$ is linearly independent modulo $(2)$. Thus $j(\Delta)\setminus\{0\}=j(\Delta')$ is linearly independent.

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    $\begingroup$ I haven't thought all of this through yet, but it's definitely necessary to start with the Borel-Tits paper (which Borel drew on heavily in writing up the results in his later work, including the added topics in his Springer GTM edition). Here is the numdam link: numdam.org/item/PMIHES_1965__27__55_0 $\endgroup$
    – Jim Humphreys
    Apr 12, 2017 at 13:34
  • $\begingroup$ I know no better proof either, and I used a variant of this in my work on relative roots for Lie algebras. The first bracket in your (2) seems to make no sense though, I guess you mean just $\sigma \in \Delta^0$. $\endgroup$
    – Torsten Schoeneberg
    Jul 23, 2017 at 18:48
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    $\begingroup$ @Torsten Schoeneberg: What I meant was a set of equations. $\endgroup$
    – Friedrich Knop
    Jul 24, 2017 at 7:53

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