Let $X$ be a possibly singular projective scheme which admits a torus $T$ action and has finitely many $T$ fixed points and one-dimensional $T-$orbits. There are many such schemes in the Grassmannian/flag variety for an algebraic group/Kac-Moody group. Then some theorems by GKM(Goresky-Kottwitz-MacPherson) allow us to compute the (equivariant)cohomology of $X$ from the moment graph of $X$, which consists of the zero and one dimensional $T-$orbits.

Based on the Bialynicki-Birula Decomposition (or equivalently on Morse Theory), we can give $X$ a "cell decomposition", and the cells are the "attracting sets" of the fixed points. My questions are: how is the information of this cell decomposition reflected in the moment graph defined above? In particular, given a $T-$fixed point $p$, how is its attracting set/cell related to the set of one-dimensional edges connected to $p$ in the moment graph for $X$? For example, can we get a bound on the dimension of this cell by counting the number of the one-dimensional edges connected to $p$ in the moment graph? Any references would be appreciated. Thanks!

  • $\begingroup$ I should add that the $\mathbb{C}^*$ required for the Bialynicki-Birula decomposition came from the torus $T$ mentioned above. $\endgroup$
    – Qiao
    Oct 31, 2017 at 15:50

1 Answer 1


The moment graph comes with an additional structure: every edge is labelled by a character of $T$. More precisely, every edge $e$ corresponds to a 1-dimensional $T$-orbit closure $C_e$. Its normalization is isomorphic to $\mathbf P^1$. An orientation of $e$ determines which fixed point in $C$ corresponds to $0$ or $\infty$, respectively. Thus, each oriented edge determines a character,namely the one with which $T$ acts on the affine line $\mathbf P^1\setminus\{\infty\}$. If $-e$ equals $e$ with the reverse orientation then $\chi_{-e}=\chi_e^{-1}$.

Now let $\lambda:\mathbf G_m\to T$ be a 1-parameter subgroup and let $X^\lambda$ be its fixed point set. If $\dim X^\lambda>0$ then $X^\lambda$ would contain one of the curves $C_e$ which means $\langle\chi_e|\lambda\rangle=0$.

So assume $\langle\chi_e|\lambda\rangle\ne0$ for all $e$. Then $X^\lambda=X^T$. Let $v\in X^T$ considered as a vertex of the moment graph. Let $e$ be an edge with outgoing orientation form $v$. Then the curve $C_e$ lies in the Bialinicki-Birula cell $X(v)$ if and only $\langle\chi_e|\lambda\rangle>0$.

With this observation it should be possible to glean a lot of information about the BB-decomposition. For example using a Noether normalization argument one can show that $X(v)$ contains at least $\dim X(v)$ $T$-stable curves. This yields an upper bound for $X(v)$. If $X$ is smooth in $v$ this becomes even an equality.


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