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 $R^+=\{e\_i-e\_j : 1\le i<j\le n\}$ 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.] 



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\}$
Unless I'm missing something obvious:


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 $e\cdot s\_{\alpha}=s\_{\alpha}$.


EDIT: By the way, the fact that rational smoothness implies smoothness for Schubert varieties in the simply laced types (and I think in *exactly* the simply laced types) is a result of Dale Peterson. See e.g. these notes: <http://www.math.ubc.ca/~carrell/SPTVGB.pdf>


