A couple of months ago, i saw a construction, that somehow looks like the construction of the tautological section of the pullback of a vector bundle to its total space, i am trying to piece it together, hope that somebody can help me:

Let $X$ be a projective scheme, say a surface over $\mathbb{C}$, let $S=Spec(R)$ be a parameter scheme. Assume we have a familiy $\mathcal{F}$ of coherent sheaves on $X\times S$, flat over $S$, let $\pi: X\times S \rightarrow S$ be the projection.

Now assume $V=\pi_{*}\mathcal{F}$ is locally free on $S$ and commutes with base change. We denote the total space of the vector bundle associated to $V$ by $T$, so $T=Spec(Sym(V^{\vee}))$, and denote by $\alpha: T \rightarrow S$ the projection.

Define $\beta$ to be the map $(id,\alpha): X\times T \rightarrow X\times S$.

Question: Does $\beta^{*}\mathcal{F}$ have a canonical (or tautological) global section?

An element of $X\times T$ is a triple $(x,s,t)$ where $t\in H^0(X_s,\mathcal{F}_s)$. Now the fiber of $\beta^{*}\mathcal{F}$ over $(x,s,t)$ should be just the fiber of $\mathcal{F}$ over $(x,s)$. So i could look at the image $t((x,s))$ of $t_{(x,s)}$ in the fiber over $(x,s)$.

Is $\sigma: X\times T \rightarrow \beta^{*}\mathcal{F}$, $(x,s,t)\mapsto t((x,s))$ a well defined global section, that is: do we have $\sigma\in \Gamma(X\times T,\beta^{*}\mathcal{F})$? Is this a canonical or tautological section? I mean, there is no other reasonable way to define the map $\sigma$. Does this make sense at all?


2 Answers 2


What you have is the fiber diagram.

$$ \begin{array}[c]{ccc} X\times T & {\stackrel{\beta}{\rightarrow}} & X\times S \\ {\scriptstyle \delta} \downarrow & & \downarrow\scriptstyle{\pi}\\ T & {\stackrel{\alpha}{\rightarrow}} & S \end{array} $$

Now, there is a natural morphism $\alpha^* V = \alpha^*\pi_*\mathcal{F}\to \delta_*\beta^*\mathcal{F}$. Compose it with the canonical section $\mathcal{O}_T\to \alpha^*V$. By adjunction, you get a section $\mathcal{O}_{X\times T}\to \beta^*\mathcal{F}$.

  • $\begingroup$ Thanks. That is much more illuminating than the explicit construction I had in mind. $\endgroup$
    – Bernie
    May 13, 2015 at 11:16

Anton has already answered, but here's a (very) slightly different explanation: for any locally free sheaf $E$ of finite rank on a scheme $X$, the 'geometric' vector bundle $V=\underline{\mathrm{Spec}}_{X}(\mathrm{Sym}(E^\vee))$ has the following functorial description: for a scheme $f\colon T\to X$ over $X$, the set $\mathrm{Hom}_X(T,V)$ is naturally in bijection with set of global sections of the locally free sheaf $f^*E$ on $T$.

In your case, $X\times T\to T$ is such a morphism, so (with Anton's notations) $\delta^*\alpha^*\pi_* F$ has a 'tautological' global section on $X\times T$, that you can see as a morphism $\mathcal{O}_{X\times T}\to \delta^*\alpha^*\pi_* F$. But now $ \delta^*\alpha^*\pi_* F=\beta^*\pi^*\pi_*F$, and composing with the adjunction morphism $\pi^*\pi_*F\to F$, you get a global section of $\beta^*F$ (the same as in Anton's answer).

  • $\begingroup$ I actually quite enjoy your perspective. $\endgroup$ May 13, 2015 at 16:50
  • $\begingroup$ It's literally the same thing you wrote, I just used the adjunction in a different step :) thanks though $\endgroup$ May 13, 2015 at 18:56

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