Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\varphi \colon X \rightarrow Y$ be a morphism of varietys, $\mathcal{F}$ a locally free sheaf of $\mathcal{O}_{Y}$-modules and $L$ the associated vector bundle. Then we can construct the inverse image sheaf $\varphi^* \mathcal{F} = \varphi^{-1} \mathcal{F} \otimes_{\varphi^{-1} \mathcal{O}_{Y}} \mathcal{O}_{X}$. Furthermore there is the so called pullback bundle $\varphi^*L = L \times_Y X$.

According to,"the pullback of bundles corresponds to the inverse image of sheaves" but how can I see this?

Thanks in advance!

share|cite|improve this question

The way I think about this (and it should be isomorphic to the way any reputable source does) is the following:

$L$ is locally equal to $\operatorname{Spec} S(\mathcal{F})$, the spectrum of the symmetric tensor algebra over $\mathcal{F}$. This is pretty easy to understand on an affine patch—you are taking formal sums of tensors of module elements $a+b\otimes c + d\otimes e\otimes f \otimes g + u\otimes w \ldots$, where you quotient out by $x\otimes y - y\otimes x$ so that it's commutative.

Since $\mathcal{F}$ is locally free, it's not too hard to see that $S(\mathcal{F})$ is locally isomorphic—over an open affine patch $\operatorname{Spec} A=U\subset Y$ small enough to trivialize $\mathcal{F}$—to $A[x_1,\ldots,x_n]$, where $n$ is the rank of $\mathcal{F}$.

Let me make that explicit. On $U$, $\mathcal{F}$ looks like the module $M=\bigoplus_{i=1}^n Ax_i$. Then $S(M)$ is exactly the free algebra over $A$ generated by $x_1, \ldots , x_n$. So $S(M) = A[x_1,\ldots,x_n]$, and we can see that it's reasonable to call $\operatorname{Spec} S(\mathcal{F})$ a "vector bundle", since locally it looks like $\mathbb{A}^n_U$.

Okay, now let's suppose that $X\to Y$ is locally given by a morphism of rings $A\to B$, where $\operatorname{Spec} B = V \subset X$. The pullback of $M$ is clearly the base change $M\otimes_A B = \bigoplus_{i=1}^n B x_i$.

But how about pulling back the bundle? Well, it's just $\mathbb{A}^n_U \times_U V = \mathbb{A}^n_V$. Or, on the level of rings, we have $S(M) \otimes_A B = S(M\otimes_A B)$. And that works at the sheaf level, too: $S(\mathcal{F}) \otimes_{\varphi^{-1} \mathcal{O}_Y} \mathcal{O}_X = S(\mathcal{F}\otimes_{\varphi^{-1} \mathcal{O}_Y} \mathcal{O}_X)$.

So this all adds up to a pretty strong dictionary (actually, an equivalence of categories) between locally free sheaves and so-called "geometric vector bundles". Actually, it's a little more general than that—if you look closely at how morphisms of sheaves lift up to the corresponding bundles, you find that as long as $\mathcal{F}$ is reflexive—that is, the natural map $\mathcal{F}\to\mathcal{F}^{\vee\vee}$ is an isomorphism—it can pretty much be identified with its bundle—the sheafy $\operatorname{Spec} S(\mathcal{F})$. You can think of these as singular bundles.

share|cite|improve this answer
Your answer made this a lot clearer for me. But there remains the following question: The pullback of the bundle locally looks like $\mathbb{A}^n_U \times _U V= Spec(S(M)) \times _{Spec(A)} Spec(B)$. To answer my question I think I should show that this is locally the same as $Spec(S(\varphi^{-1}{F} \otimes _{\varphi^{-1} \mathcal{O}_Y} \mathcal{O}_X))$ but I don't know how to do this. – claudi Oct 18 '13 at 11:36
@claudi It's really about the simpler-looking $S(M)\otimes_A B = S(M\otimes_A B)$. You could do this with universal properties, or you could look at $S(M)$, as a module, as the direct sum $\bigoplus_m M^{\otimes m}$. – Slade Oct 18 '13 at 17:37

$Sh(X)\leftrightarrows Et(X)$ is the deep reason; for example in "Sheaves in Geometry and Logic" the inverse image sheaf is defined as the sheaf associated to the pullback of the bundle which corresponds to the sheaf you're considering.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.