# In Type $A$, if the Bruhat graph of an element $w$ in the Weyl group is regular, then to show that $l(w)=$ # $\{\alpha \in R^+| s_{\alpha} \le w\}$

I am trying to prove that for type $$A$$ , rational smoothness of Schubert varieties implies smoothness.

So suppose we are in Type $$A_{n-1}$$, so let $$G=Sl(n,\mathbb C)$$, $$B=$$ the group of upper triangular matrices in $$G$$ , $$T=$$ group of diagonal matrices in $$G$$. $$W=N_G(T)/T \cong \mathfrak S_n$$. Let $$R=\{e_i -e_j : 1\le i, j\le n, i\ne j\}$$ be our root system and $$R^+=\{e_i-e_j : 1\le i be positive root system.

Let $$\le$$ be the Bruhat order on $$W$$.

For $$w\in W$$ , let $$\Gamma(w)$$ be the Bruhat graph of $$w$$, whose vertex set is $$\{u \in W : u\le w\}$$ and between any two vertices $$x$$ and $$y$$, there is a edge if and only if $$x=ys$$ for some reflection $$s$$ (not necessarily simple).

Now suppose it is given that every vertex of the Bruhat graph $$\Gamma (w)$$ has degree equal to $$l(w)$$ (i.e. equivalently the Schubert variety $$X(w)$$ is rationally smooth ). Then how to show that

$$l(w)=$$ # $$\{\alpha \in R^+| s_{\alpha} \le w\}$$ ?

[Note that if I can show my claim then I am done because $$l(w)=\dim X(w)$$ and dimension of the tangent space of $$X(w)$$ at $$e_{id}$$ is $$\dim T(w,e_{id})=$$# $$\{\alpha \in R^+| s_{\alpha} \le w\}$$, so the claim would imply $$X(w)$$ is smooth at $$e_{id}$$ , hence smooth.]

Consider the degree of the identity element $$e$$ in $$\Gamma(w)$$. On the one hand, by the assumption you've made it is $$\ell(w)$$. On the other hand, it is clearly equal to the number of reflections $$s_{\alpha}$$ less than $$w$$, because the only things $$e$$ will be connected to are of the form $$e\cdot s_{\alpha}=s_{\alpha}$$.
• Thanks a lot ! And yes I know that result ... I am only concerned with the classical types $A,B,C,D$ in which case, rational-smooth ness implies smoothness only for Type A and D... the Type A case is easier which is what I was trying to prove by hand first – user102248 May 7 '19 at 22:05