# Grassmannian a Fano manifold?

I am interested in answering the following: Is it true that the Grassmannian $G(k,\mathbb{C}^n)$ is a Fano manifold? How can I see if its anticanonical bundle is ample?

• You can see this by considering the Plücker embedding. Commented Mar 21, 2015 at 21:36

The main tool is the canonical exact sequence of vector bundles on $G=G(k,\mathbb{C}^n)$:$$0\to S\to G(k,\mathbb{C}^n)\times \mathbb C^n\to Q\to 0 \quad (\bigstar)$$ where $S$ is the rank $k$ tautological bundle, and Q is the rank $(n-k)$ quotient bundle.
From the expression $T=\operatorname {Hom}(S,Q)=S^*\otimes Q\;$ of the tangent bundle $T=T_G$ to $G$ we get the equality for the anticanonical bundle (I write $\wedge^{top} E :=\wedge^r E$ for a vector bundle of rank $r$) : $$\wedge ^{top}T=(\wedge ^{top} S^*)^{\otimes (n-k)}\otimes (\wedge ^{top} Q)^{\otimes k} \quad (\bigstar \bigstar)$$ Now $(\bigstar)$ yields $\wedge ^{top} Q=\wedge ^{top} S^*$ and thus the expression $(\bigstar \bigstar)$ for the anticanonical bundle of $G$ becomes $$\wedge ^{top}T=(\wedge ^{top} S^*)^{\otimes (n-k)}\otimes (\wedge ^{top} S^*)^{\otimes k}=(\wedge ^{top} S^*)^{\otimes n} \quad (\bigstar \bigstar \bigstar)$$ Now in the Plücker embedding $$j:G\hookrightarrow \mathbb P(\wedge^k \mathbb C^n)=:\mathbb P$$ we have $j^*\mathcal O_\mathbb P (1) =\wedge ^{top} S^*$, which is thus ample and thus a fortiori the anticanonical bundle of $G$ described in $(\bigstar \bigstar \bigstar)$ is ample since it is the $n$-th power of that ample bundle: $$\operatorname {Anticanonical bundle}=\wedge ^{top}T= (j^*\mathcal O_{\mathbb P} (1))^{\otimes n}=\operatorname {ample}$$ For those knowledgeable about the Schubert calculus, let me mention that $\wedge ^{top} S^*$ is the line bundle associated to the divisor $\Sigma_1\subset G(k,\mathbb C^n)$ . The canonical divisor can thus be written $K_G=-n\Sigma_1$.