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Let $V$ be a finite dimensional vector space over $\mathbb{C}$, denote $V^{\times} = V -0$ and let $\pi : V^{\times} \rightarrow \mathbb{P}(V)$ be the quotient by the natural action of $\mathbb{C}^{\times}$ on $V$. If $M$ is a coherent $D$-module on $\mathbb{P}(V)$, then the action of $\mathbb{C}^{\times}$ on the global section $\Gamma(V^{\times}, \pi^*M)$ gives a weight decomposition: $$ \Gamma(V^{\times}, \pi^*M) = \bigoplus_{l \in \mathbb{Z}} \Gamma(V^{\times}, \pi^*M)^l,$$ where $z \in \mathbb{C}^{\times}$ acts by multiplication by $z^l$ on $\Gamma(V^{\times}, \pi^*M)^l$. Now, let $E$ be the Euler vector field on $V^{\times}$, that is: $$E = \sum_{i=1}^{n} x_i \frac{\partial}{\partial x_i},$$ for $x_i$ a base of $V$. Since $\pi^*M$ is a $D$-module, we have a natural action of $E$ on $\Gamma(V^{\times}, \pi^*M)$, and $\Gamma(V^{\times}, \pi^*M)^l$ is the eigenspace with eigenvalue $l$ for the action of $E$ on $\Gamma(V^{\times}, \pi^*M)$.

This being said, I was vondering if something similar is true for Grassmannians. Namely let $X = \mathbb{G}(k,V)$ be the Grassmannians of $k$ dimensional vector subspaces in $V$. Let $X^{\times}$ be the affine cone over $X$ minus $0$ and $\pi : X^{\times} \rightarrow X$ be the quotient by the action of $\mathbb{C}^{\times}$ on $X^{\times}$. Let $M$ be a coherent $D$-module on $X$. We still have a decomposition: $$\Gamma(X^{\times}, \pi^*M) = \bigoplus_{l \in \mathbb{Z}} \Gamma(X^{\times}, \pi^*M)^l,$$ where $z \in \mathbb{C}^{\times}$ acts by multiplication by $z^l$ on $\Gamma(X^{\times}, \pi^*M)^l$.

My question is the following : is there a globally defined differential operator on $X^{\times}$ (say $E$) such that the eigenspaces for the action of $E$ on $\Gamma(X^{\times}, \pi^*M)$ are precisely the $\Gamma(X^{\times}, \pi^*M)^l$ ? (I was hoping this because the tangent bundle to the Grassmannian is globally generated)

If not, is there at least an analogous phenomenon for the Grassmannian?

Thanks a lot!

share|cite|improve this question
If you have a $\mathbf{C}^\times$-bundle $L\rightarrow X$, you get an action map $\mathrm{Lie}\mathbf{C}^\times\rightarrow H^0(L, T_L)$. The image of $z\partial_z$ under this map is the required vector field. Isn't that what you're asking? – Pavel Safronov Feb 9 '13 at 0:59
Dear Pavel, thanks for your answer! Since I am just a beginner in Lie theory, could just explain why the eigenspaces of the vector field you defined are precisely the $\Gamma(X^{\times}, \pi^*M)^l$? – Johan Feb 9 '13 at 12:36
The pullback locally looks like $\pi^{-1}M\otimes\mathcal{O}_{\mathbf{C}^\times}$. The decomposition $\mathcal{O}_{\mathbf{C}^\times}=\oplus\mathbf{C}\cdot z^l$ is precisely the eigenspace decomposition for $z\partial_z$. There is also an invariant way to write the pullback as $\pi^{-1}M\otimes_{\pi^{-1}\mathcal{O}_X}\mathcal{O}_L$; observe that $E$ acts trivially on $\pi^{-1}\mathcal{O}_X$. – Pavel Safronov Feb 11 '13 at 16:37

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