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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

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    $\begingroup$ 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. $\endgroup$ Commented Jun 16, 2021 at 14:08
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    $\begingroup$ @FrancescoPolizzi thank you for the answer! But do you know what are, from a geometric description/point of view, elements in $H^0(T_X(d)) \setminus H^0(T_X)$? $\endgroup$
    – gigi
    Commented Jun 16, 2021 at 14:15
  • $\begingroup$ 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 $Y \supset\operatorname{supp}\mathcal F$. $\endgroup$ Commented Jun 17, 2021 at 4:36

2 Answers 2

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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?

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  • $\begingroup$ 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? $\endgroup$
    – gigi
    Commented Jun 16, 2021 at 15:23
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    $\begingroup$ @gigi I guess these would be of the generated by the Euler vector field $\sum x_i\partial/\partial x_i$. $\endgroup$ Commented Jun 16, 2021 at 15:30
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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 $G1$.

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    $\begingroup$ 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. $\endgroup$
    – gigi
    Commented Jun 16, 2021 at 14:47

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