# Where does the half sum of positive roots live?

Definitions: Let $$G$$ be a compact, connected, semi-simple (defined below) Lie group with maximal torus $$T$$, Weyl group $$W:=N(T)/T$$ and lie algebra $$\mathfrak{g}$$. Let $$\Lambda$$ be the dual of $$T$$, and let $$\Lambda_{root}$$ be the sub-lattice spanned by the weights ($$\Phi$$) of the adjoint representation of $$T$$ on $$\mathfrak{g}\otimes \mathbb{C}$$. Semi-simplicity means the index $$[\Lambda:\Lambda_{root}]$$ is finite. Give $$\Lambda \otimes \mathbb{R}$$ an inner product invariant under $$W$$ and split $$\Lambda \otimes \mathbb{R}$$ into two half-spaces to define the set of positive roots $$\Phi^+$$.

Question: The book I'm reading (Bump, Lie Groups, 2nd edition) claims that $$\rho:=\frac{1}{2}\sum_{\Phi^+}\alpha$$ is in $$\Lambda$$ (First line of the proof of prop $$22.2$$). How do I see this? This fact is not obvious to me and I can't find the part in the book where he's proved it.

• That's a very unusual definition of semisimple. -- Can you give a more formal definition of $\Lambda$ here ("dual" can mean many things: What Hom's to what space?), as that seems to be the crucial point. Namely, if $\Lambda=\Lambda_{root}$ can happen (candidate: groups of adjoint type, but that depends on the exact definition of $\Lambda$), then as soon as $\rho \notin \Lambda_{root}$ (e.g. type $A_3$ or $B_2$) we have counterexamples (e.g. $PSU(4)$ or $SO(5)$). – Torsten Schoeneberg Feb 6 at 0:58
• $\Lambda$ is the character group of $T$, ie. continuous group homomorphisms to the circle. The definition of semi-simple is from the linked book, at the start of chapter $22$. If there are counterexamples, can you please address my confusion regarding prop $22.2$? – Idi Amin Feb 6 at 14:01
• Your confusion is justified, I don't think the theorem is correct as stated. In general, one would need to assume that $G$ is simply connected, i.e replace $\Lambda$ by $\tilde \Lambda$ as def. on p. 166. Three sources which explicitly mention that condition: web.stanford.edu/~tonyfeng/222.pdf p.109, Knapp's "Representation Theory of Semisimple Groups" p.105, Procesi's "Lie Groups" p.469. However, understanding the representations of the simply connected cover includes understanding it for general $G$ by the procedure outlined on p.147. – Torsten Schoeneberg Feb 7 at 20:07
• @TorstenSchoeneberg The definition of semi-simple in terms of the index of the root lattice seems to be standard (in the context of the group already being assumed reductive for the root system to be defined in the first place). – Tobias Kildetoft Feb 7 at 20:32
• Thanks for going through the reference, Torsten. So I guess I could read the rest of the chapter assuming $G$ is simply connected. I'm not sure what to do about this question though; whether to accept David's answer or to wait for Torsten/Tobias to put down their comments as an answer. – Idi Amin Feb 7 at 20:50

This follows from the fact that if $$\alpha\in\Delta=\{\alpha_1,\ldots,\alpha_n\}$$ is a simple root and $$s_\alpha$$ is the reflection $$s_\alpha(\alpha)=-\alpha$$ and $$s_\alpha(\beta)=\beta$$ for $$\beta\in\alpha^\perp$$, then $$s_\alpha$$ permutes $$\Phi^+\backslash\{\alpha\}$$. Therefore, $$s_\alpha(\rho)=\rho-\alpha$$ which shows that $$(\rho,\alpha)=(\alpha,\alpha)/2$$.
Now, let $$\omega_1,\ldots,\omega_n$$ be the fundamental dominant weights (so $$2(\omega_i,\alpha_j)/(\alpha_j,\alpha_j)=\delta_{ij}$$). Then, writing $$\rho=\sum_ia_i\omega_i$$ we have $$(\rho,\alpha_j)=\sum_ia_i(\omega_i,\alpha_j)=a_j(\alpha_j,\alpha_j)/2.$$ Now, solving for $$\alpha_j$$ one easily deduces the formula $$\rho=\sum_i\frac{2(\rho,\alpha_i)}{(\alpha_i,\alpha_i)}\omega_i=\sum_i\omega_i.$$ Therefore, $$\rho$$ is a sum of the fundamental weights.
As pointed out by Tobias Kildetoft in the comments, the fundamental weights belong to the weight lattice provided $$G$$ is simply connected.
• Hi! Fundamental weights are not necessarily in $\Lambda$, are they? (See the third paragraph at the start of chapter $21$ in the 2nd edition of Bump's book) – Idi Amin Feb 4 at 17:49
• Yes they are. The fundamental weights are the elements of $\mathrm{Hom}(T,\mathbb{Z})$ given my $\omega_i(H_\alpha)=(\omega_i,\alpha)$, $\alpha\in\Phi^+$. They obviously form a basis of $\Lambda$. – David Hill Feb 4 at 18:17
• You mean $Hom(T, S^1)$? Doesn't $s_{\alpha}(\rho)=\rho-\alpha$ imply $(\rho,\alpha)=(\alpha,\alpha)/2$? What is $H_{\alpha}$? – Idi Amin Feb 4 at 19:18
• I don't think I mean $Hom(T, S^1)$. You are right about $(\rho, \alpha)$. I used the wrong bilinear form and will update later. – David Hill Feb 4 at 21:30
• The argument using fundamental weights requires that the group is simply connected. Without this assumption the claim can fail (as I recall, it should fail for a suitable $PGL_n$ in the algebraic group case, though this may need tweaking in the setting here). – Tobias Kildetoft Feb 4 at 22:14