# Geometric meaning of coherent sheaves $\mathcal{F} \otimes \mathcal{O}_{\mathbb{P}^n}(d)$ over $\mathbb{P}^n$

Maybe it sounds like a silly question but I'm not able to figure out in my head the geometric meaning of "twisting" vector bundles (or more generaly coherent sheaf) over $$\mathbb{P}^n$$.

I explain myself better: if I take $$X=\mathbb{P}^n$$ and then I consider a subscheme $$Y \subset X$$, then to $$Y$$ corresponds an ideal (saturated if you want) $$I_Y=(F_1,\dots,F_r) \subset \mathbb{C}[x_0,\dots,x_n]$$ Sheafifying I can view $$\mathcal{I}_Y$$ as a coherent $$\mathcal{O}_X-$$module. I can also consider its twist by $$\mathcal{O}_X(d)$$, i.e. $$\mathcal{I} \otimes \mathcal{O}_X(d)=\mathcal{I}(d)$$ I have a very clear geometric meaning of sections of $$\mathcal{I}(d)$$, i.e. $$H^0(\mathcal{I}(d))$$ corresponds to hypersurfaces $$F \in \mathbb{P}^n$$ of degree $$d$$ containing $$Y$$. Now, if for example I take the Tangent sheaf $$\mathcal{T}_X$$, it is also a coherent $$\mathcal{O}_X-$$module, and geometrically $$H^0(\mathcal{T}_X)$$ is the space of vector fields over $$\mathbb{P}^n$$. Now, if I consider the twist $$\mathcal{T}_X \otimes \mathcal{O}_X(d)=\mathcal{T}_X(d)$$ what is the geometric meaning of the sections in $$H^0(\mathcal{T}_X(d))$$? Are they a sort of vector fields with special properties?

And more generally if I have a nice geometric description of sections of a coherent sheaf $$H^0(\mathcal{F})$$, how can I find a geometric interpretation of the sections of $$H^0(\mathcal{F}(d))$$? Thanks in advance for the help

• Well, if $d \geq 0$ in general we have $h^0(T_X(d)) \geq h^0(T_X)$, so the sections of $T_X(d)$ are not only vector fields. Since $H^0(T_X) \subseteq H^0(T_X(d))$, vector fields are contained here. Commented Jun 16, 2021 at 14:08
• @FrancescoPolizzi thank you for the answer! But do you know what are, from a geometric description/point of view, elements in $H^0(T_X(d)) \setminus H^0(T_X)$?
– gigi
Commented Jun 16, 2021 at 14:15
• Positive twisting can make rational sections regular, so one interpretation is that global sections of $\mathcal F(d)$ have at least something to do with rational sections of $\mathcal F$ that blow up in a controlled way along a degree $d$ divisor $Y$. If $\mathcal F$ is a vector bundle, then from $0 \to \mathcal F \to \mathcal F(d) \to \mathcal F(d)|_Y \to 0$ we have $H^0(\mathcal F) \subset H^0(\mathcal F(d))$ which makes this interpretation a bit more literal. OTOH if $\mathcal F$ is a torsion sheaf, one could perhaps leverage this idea by choosing $Y \supset\operatorname{supp}\mathcal F$. Commented Jun 17, 2021 at 4:36

Let $$x_0,\ldots, x_n$$ be homogeneous coordinates, then a section of $$H^0(\mathbb{P}^n,\mathcal{T}_{\mathbb{P}^n}(d))$$ can be expanded as a sum $$\sum_i f_i(x_0,\ldots, x_n) \frac{\partial}{\partial x_i}$$ where $$f_i$$ are homogenous polynomials of degree $$d$$. This follows more or less immediately from the Euler sequence $$0\to \mathcal{O}\to \bigoplus_0^n \mathcal{O}(1)\to \mathcal{T}\to 0$$ after twisting.

Does this help?

• I think that by Euler sequence $$0 \rightarrow \mathcal{O}_{\mathbb{P}^n}(-1) \rightarrow \mathcal{O}_{\mathbb{P}^n}^{n+1} \rightarrow \mathcal{T} \rightarrow 0$$ after taking the twist by $\mathcal{O}(d)$ you have to take into the account the Kernel coming from $\mathcal{O}_{\mathbb{P}^n}(d-1) \rightarrow \mathcal{O}(d)_{\mathbb{P}^n}^{n+1}$ that will be a $\binom{n+d-1}{n}$ dimensional vector space of relations among the homogenenous polynomials $(f_0,\dots,f_n)$. Do you know how to express these relations and what they mean?
– gigi
Commented Jun 16, 2021 at 15:23
• @gigi I guess these would be of the generated by the Euler vector field $\sum x_i\partial/\partial x_i$. Commented Jun 16, 2021 at 15:30

Question: "what is the geometric meaning of the sections in $$H^0(T_X(d))$$? Are they a sort of vector fields with special properties?"

Answer: Let $$C:=\mathbb{P}^1:=\mathbb{P}(V^*)$$ with $$V:=k\{e_0,e_1\}$$. If you want to relate the invertible sheaf $$L:=\mathcal{O}_{C}(d)$$ ($$d \geq 0$$) to geometry, maybe a good idea is to consider the $$d$$-uple embedding

$$\phi_d: C \rightarrow \mathbb{P}^d:=X.$$

It follows $$\phi_d^*(L) \cong \mathcal{O}_{C}(d)$$. The tangent sheaf $$\mathcal{T}_C \cong \mathcal{O}_{C}(2)$$ on $$C$$ hence you are interested in the tensor product

$$T_C(d):=T_C \otimes \phi_d^*(L)$$

and the global sections

$$H^0(C, T_C(d)) \cong H^0(C, \mathcal{O}_C(d+2)) \cong Sym^{d+2}(V^*)$$

where $$V^*:=k\{x_0,x_1\}$$. Maybe you can relate these global sections to the embedding $$\phi_d$$. There is a well defined multiplication map

$$d: Sym^2(V^*)\otimes Sym^d(V^*) \rightarrow Sym^{d+2}(V^*)$$

with $$d(u\otimes v):=uv$$.

If $$I_d \subseteq \mathcal{O}_{X}$$ is the ideal sheaf of $$C$$ in the $$d$$-uple embedding there is an exact sequence

$$C1.\text{ }0 \rightarrow I_d/I_d^2 \rightarrow \phi_d^*(\Omega^1_{X/k}) \rightarrow \Omega^1_{C/k} \rightarrow 0,$$

hence the cotangent sheaf $$\Omega^1_{C/k}\cong \mathcal{O}(-2)$$ may be constructed as a quotient of the pull back of the cotangent sheaf of $$X$$. This is the dual of the tangent mapping wrto $$\phi_d$$. If you want to relate invertible sheaves and tangent bundles/cotangent bundles to "geometry" you must study the relationship between invertible sheaves and maps to projective space and the sequence $$C1$$. Tensor the sequence $$C1$$ and dualize to get the sequence

$$0 \rightarrow T_C \otimes \mathcal{O}(d) \rightarrow \phi_d^*T_X \otimes \mathcal{O}(d) \rightarrow Hom(I_d/I_d^2\otimes \mathcal{O}(-d), \mathcal{O}) \rightarrow 0$$

and take global sections to get

$$G1.\text{ } 0\rightarrow H^0(C, T_C \otimes \mathcal{O}(d)) \rightarrow H^0(C, \phi_d^*T_X \otimes \mathcal{O}(d)) \rightarrow H^0(C, Hom(I_d/I_d^2\otimes \mathcal{O}(-d) , \mathcal{O}))\rightarrow \cdots$$

The above sequence relates the global sections of $$T_C\otimes \mathcal{O}(d)$$ to the embedding $$\phi_d$$. You should identify the image of the left map as a subspace of the vector space of global sections of $$\phi_d^*T_X \otimes \mathcal{O}(d)$$. Locally for maps of rings $$k \rightarrow A \rightarrow B$$ there is the tangent sequence

$$0 \rightarrow Der_A(B) \rightarrow Der_k(B) \rightarrow B\otimes_A Der_k(A),$$

and when $$B:=A/I$$ you get the sequence

$$R1.\text{ }0 \rightarrow Der_k(A/I) \rightarrow A/I \otimes_A Der_k(A) \rightarrow Hom_{A/I}(I/I^2, A/I) \rightarrow \cdots .$$

If $$Y:=Spec(A/I) \subseteq X:=Spec(A)$$ it follows the sequence $$R1$$ relates vector fields on $$Y$$ to vector fields on $$X$$ "parallel" to $$Y \subseteq X$$. The module $$A/I \otimes_A Der_k(A)$$ is the restriction of $$Der_k(A)$$ to the closed subscheme $$Y:=V(I)\cong Spec(A/I)$$.

Note: In the case of the projective line, when you pull back the cotangent bundle you get a decomposition

$$\phi_d^*(\Omega^1_{X/k}) \cong \oplus_i \mathcal{O}_C(d_i)$$

into direct sums of invertible sheaves. There is the notion k-very ample line bundle that may be of interest. Note also that when you have a closed embedding $$\phi_d:\mathbb{P}^1 \subseteq \mathbb{P}^d$$, you may consider global sections of $$T_{\mathbb{P}^d}$$ - global vector fields $$\partial$$ on $$\mathbb{P}^d$$. The global vector fields $$\partial$$ that are parallel to $$\mathbb{P}^1$$ in the embedding $$\phi_d$$, restrict to vector fields on $$\mathbb{P}^1$$. You should relate this to the sequence $$G1$$.

• Thank you very much for your answer! My problem, sorry for not having specified it, is with $\mathbb{P}^n$ for which $n \geq 2$. In the case of $\mathbb{P}^1$, as you have observed, the tangent bundle is a line bundle itself, so everything is deducible from the "geometry of divisors". Where $n \geq 2$ then the tangent bundle is no more a line bundle and so I'm not able to see how to relate its twist with, for example, properties of vector fields.
– gigi
Commented Jun 16, 2021 at 14:47