The Lie algebra $\mathfrak{sl}_n $ has many special features which are not shared by other simple Lie algebras, for example all of its fundamental representations are minuscule.

I recently discovered another curious fact.
The number of positive roots of $ \mathfrak{sl}_n $ is the same as the dimension of $ Sym^2 \mathfrak h$ --- both are $ \binom{n}{2} $. All other simple Lie algebras have more positive roots. Is this fact is connected to the special properties enjoyed by $\mathfrak{sl}_n$? Is there some a priori reason to expect this equality?

I came upon this fact because I was thinking about the following linear map (here $\mathfrak g $ is any semisimple Lie algebra and $ \mathfrak h $ is its Cartan subalgebra): \begin{align*} Sym^2 \mathfrak h &\rightarrow U \mathfrak g \\ xy &\mapsto \sum_{\alpha \in \Delta_+} \langle \alpha, x \rangle \langle \alpha, y \rangle E_\alpha F_\alpha \end{align*} The set $ \{ E_\alpha F_\alpha \}_{\alpha \in \Delta_+} $ in $ U \mathfrak g $ is linearly independent and I was hoping that this map would be an isomorphism onto its span, but this can only be true for $ \mathfrak{sl}_n $.

Has anyone seen this map before?

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    $\begingroup$ I have not seen this map before. Up to some rescaling, I think it is something like $[xy, R]$, where $R$ is the R-matrix and the commutator takes place in $U\mathfrak g \otimes U\mathfrak g$? $\endgroup$ – Theo Johnson-Freyd May 13 '17 at 4:26
  • $\begingroup$ Interesting. If I compute correctly, the map sends $(e_{ii}-e_{jj})^2$ to $e_{ij}e_{ji}$, so it must be as claimed. $\endgroup$ – მამუკა ჯიბლაძე May 13 '17 at 15:25
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    $\begingroup$ You can identify $\operatorname{Sym}^2(\mathfrak{h})$ with the space of symmetric bilinear forms on $\mathfrak{h}$. For each root $\alpha$, you can define a symmetric bilinear form $\varphi_\alpha$ on $\mathfrak{h}$ via $\varphi_{\alpha}(h, h') = \alpha(h)\alpha(h')$ for all $h, h' \in \mathfrak{h}$. Then it seems in the case of $\mathfrak{sl}_n$ that $\{ \varphi_\alpha : \alpha \text { positive root} \}$ is linearly independent in $\operatorname{Sym}^2(\mathfrak{h})$. ... $\endgroup$ – spin May 13 '17 at 15:27
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    $\begingroup$ I wonder if the same is true for other simple types? That would at least show that the dimension of $\operatorname{Sym}^2(\mathfrak{h})$ is $\geq$ number of positive roots, then maybe there is some reason why equality holds only for type $A$? $\endgroup$ – spin May 13 '17 at 15:27
  • $\begingroup$ @spin I believe for all simples, $\sum_\alpha\varphi_\alpha$ is nondegenerate (and is, up to a scalar multiple, the only Weyl-group-invariant symmetric bilinear form on $\mathfrak h$), although I do not quite understand whether this helps in any way to answer your question $\endgroup$ – მამუკა ჯიბლაძე May 15 '17 at 5:37

I will mostly answer spin's approach and question here, but it is rather long to write as a comment only. I am not sure if it will answer the original OP's question, but it does seem like a good step.

Given a positive root $\alpha$, we associate to it, we associate to it (following spin's comment) a symmetric bilinear form $\varphi_\alpha$ on $\mathfrak{h}$, defined by:

$\varphi_\alpha(h,h') = \alpha(h)\alpha(h')$

for all $h,h'\in \mathfrak{h}$.

We claim that the set $\{\varphi_\alpha; \alpha \text{ is a positive root}\}$ spans $\operatorname{Sym}^2(\mathfrak{h})$ for simple algebras. This is equivalent to showing that any real homogeneous quadratic form $B(-,-)$ on $\mathfrak{h}^*$ which vanishes identically on the set of positive roots must vanish identically.

Lemma: if $V$ is a real vector space, with a symmetric bilinear form $B(-,-)$ on $V$, and if $v_1, v_2\in V$ are linearly independent, and suppose further that $B(v_i,v_i) = 0$ for $i=1,2$ and that $B(v_1+cv_2, v_1+cv_2)=0$ for some real nonzero constant $c$, then $B(-,-)$ must vanish identically on the span of $v_1$ and $v_2$.

The proof of the lemma is easy. Indeed it implies in particular that the $v_i$ are null, and that $B(v_1,v_2) = 0$.

We now apply this lemma inductively. We start on one end of the Dynkin diagram, and apply the lemma to $v_1$ being the first simple root, and $v_2$ being the second simple root, which is connected to the first simple root by either a simple, a double or triple edge. Applying reflection to $v_2$ about the hyperplane orthogonal to $v_1$, we get the required third vector of the form either $v_1+cv_2$ or $v_2+cv_1$ for some $c>0$, which is also a positive root. We thus get that $B(-,-)$ vanishes identically on the span of the first 2 simple roots, and then apply the lemma again to the second simple root, and a new simple root connected to the second simple root by a simple, double or triple edge, and so on.

This would show that $B(-,-) = 0$ identically, thus proving the claim.

Hence this would show that the number of positive roots is always greater or equal to the dimension of $\operatorname{Sym}^2(\mathfrak{h})$, answering spin's question at least. The proof is conceptual and does not rely on the classification result. I am not sure if it really answers the OP's original question though.

Edit 1: building up on my previous argument, we would like to show, in order to answer the OP's question, that if the $\varphi_\alpha$'s are linearly independent over $\mathbb{R}$ in $\operatorname{Sym}^2(\mathfrak{h})$, then the root system is of the A type. Thus we have to rule out the existence of multiple edges, and rule out the existence of Dynkin subdiagrams which are isomorphic to that of $SO(8)$ (here, note that I am unfortunately relying on the classification result). So let us assume that the $\varphi_\alpha$ are linearly independent. Then for any positive root $\alpha$, there exists a symmetric bilinear form $B(-,-)$ on $\mathfrak{h}^*$, such that $B(\beta,\beta) = 0$ for any positive root $\beta \neq \alpha$ and $B(\alpha,\alpha) \neq 0$. By restricting to a rank 2 Dynkin subdiagram, corresponding to 2 simple roots connected by a multiple edge, one can remove one positive root spanned by these 2 simple roots, and still get at least 3 positive roots satisfying the lemma above, so that $B$ must vanish identically on the span of these 2 simple roots, thus leading to a contradiction.

It remains to rule out the existence of rank 4 Dynkin subdiagrams isomorphic to that of $SO(8)$ under the linear independence assumption on the $\varphi_\alpha$. This can be done explicitly by removing a specific positive root, and showing that if $B$ vanishes on all other positive roots, then it must be 0 on this rank 4 subspace of $\mathfrak{h}^*$, thus leading to a contradiction too. Alternatively, one can simply count the dimensions in this case: $SO(8)$ has 12 positive roots, while $\operatorname{Sym}^2(\mathfrak{h})$ has dimension 10, thus ruling out the existence of such Dynkin subdiagrams under the linear independence hypothesis on the $\varphi_\alpha$.

Remark: the first part of my post does not rely on the classification result, while edit 1 does unfortunately rely on at least part of the classification result. Edit 1 consists in ruling out multiple edges and Dynkin subdiagrams isomorphic to that of $SO(8)$. I hope that over all, my post answers the OP's question (it could use some editing though).


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