# The image of a base of absolute roots is a base of relative roots

$G$ 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 $_k \Phi$.

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 $_k \Delta \cup \{0\}$.

• 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 Apr 12, 2017 at 3:25
• 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
– D_S
Apr 12, 2017 at 3:30
• 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. Apr 12, 2017 at 4:05
• 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. Apr 12, 2017 at 4:53
• 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... Apr 13, 2017 at 2:18

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.
• 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 $\sigma \in \Delta^0$. Jul 23, 2017 at 18:48