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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.]

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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

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  • $\begingroup$ 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 $\endgroup$ – user102248 May 7 '19 at 22:05

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