# Cohomology Class of a Subvariety

I'm working on question 7.4 of Chapter III.7 in Hartshorne's Algebraic Geometry. The question is about the cohomology class of a subvariety.

The setup is as follows: $X$ is an $n$-dimensional non-singular projective variety over an algebraically closed field $k$. $Y\subset X$ is a non-singular subvariety of codimension $p$. We have the standard map $\Omega_X\otimes \mathcal{O}_Y \rightarrow \Omega_Y$ from which we can deduce a map $\Omega_X^{n-p} \rightarrow \Omega_Y^{n-p}$, and this in turn yields a map on cohomology $H^{n-p}(X, \Omega_X^{n-p}) \rightarrow H^{n-p}(Y,\Omega_Y^{n-p})$. Now $Y$ is $(n-p)$-dimensional so $\Omega_Y^{n-p} = \omega_Y$ and we have the trace map $H^{n-p}(Y,\Omega_Y^{n-p} = \omega_Y) \rightarrow k$. Composing with this trace map we have $\varphi_Y : H^{n-p}(X,\Omega_X^{n-p})\rightarrow k$. Now, since by Serre Duality $H^p(X,\Omega_X^p) \cong H^{n-p}(X, \Omega_X^{n-p})^{\lor}$, $\varphi_Y$ corresponds to an element $\eta(Y) \in H^p(X,\Omega_X^p)$ which we call the cohomology class of $Y$.

Part (a) of the problem asks to show that if $P\in X$ is a closed point, then $t_X(\eta(P)) = 1$, where $t_X$ is the trace map on $\omega_X$. Now, in the case of a point, the map $\Omega_X^{n-p} \rightarrow \Omega_Y^{n-p}$ is simply the map $\mathcal{O}_X \cong \Omega_X^0 \rightarrow \Omega_P^0\cong k_P$ defined by $f \mapsto f(P)$ which, when composed with the trace map on $H^0(P,\omega_P)$, yields $f \mapsto t_P(f(P))$. This map should correspond to an element $\eta(P) \in H^n(X,\Omega_X^n = \omega_X)$ as indicated above.

My problem is, I don't know how to get my hands on $\eta(P)$. All I know is that it should exist. Moreover, I thought that $H^n(X, \omega_X) \rightarrow k$ is only a map of $k$-vector spaces. Wouldn't I need some morphism respecting ring structures somewhere in order to detect $1 \in k$ and be able to show that $t_X(\eta(P)) = 1$ here? Hartshorne himself says that, except in the case of curves, we can't really write down explicitly what the trace map is because we don't know. All we know is that it exists.

• In the case of a closed point we have $n=p$, so the map inducing $\eta(P)$ is $H^0(X,\mathcal O_X) \to H^0(P,\mathcal O_P) \to k$. So the map should be the natural isomorphism from global sections of the structure sheaf to $k$, which does "see" the ring structure. – Tabes Bridges May 22 '16 at 0:24
• I see what you're saying, but wouldn't that be implying that the map $f \mapsto t_P(f(P))$ sees the ring structure rather than the map $t_X$, which is the one I want to have $t_X(\eta(P)) = 1$? – user341315 May 22 '16 at 0:30
• Sorry, what I'm trying to say (although the more I think about it, the more slippery the idea becomes) is that $\eta(P)$ is the image, under the duality isomorphism, of a particularly natural map $H^0 \to k$, which I want to say is the $1$ element in $(H^0)^\vee$. I say that the map is particularly natural because (unlike in the cases of higher cohomology and higher powers of $\Omega$) it makes sense to talk about elements of the domain (which are simply constant functions!) taking particular values in $k$. So even though $H^n$ is somewhat abstract, the other side of the isomorphism is clear. – Tabes Bridges May 22 '16 at 0:57
• Tabes, thanks for your clarification. If I'm understanding you correctly, you want to say that the composed map $H^0(X,\Omega_X^0 = \mathcal{O}_X) \to H^0(P,\Omega_P^0 \cong \mathcal{O}_P) \to k$ (which probably looks like $f \mapsto t_P(f(P))$) sends $"1" \in H^0(X,\mathcal{O}_X)$ to $1 \in k$. So it seems there like you want to say that $t_P(1) = 1$. Now saying where $"1" \in H^0(X,\mathcal{O}_X)$ gets sent determines the map, so you want to call that particular map... – user341315 May 22 '16 at 13:09
• ...the "1 element" in $H^0(X,\mathcal{O}_X)'$ which, via the duality isomorphism should get sent to the element $\eta(P) \in H^n(X,\omega_X)$. But, even assuming we accept everything up to now, I still don't even see how $t_X(\eta(P))$ should equal $1$. Presumably we want it to equal $t_P(1)$ at least? But I don't even see that. – user341315 May 22 '16 at 13:09

Well, I think you have to make a choice for $t_{P}$. But apparently you do not need to make a choice for $t_X$. This happens because the role of $t_X$ in the duality isomorphism gets cancelled when you evaluate the class using $t_X(. )$.

At first, for a closed immersion $j: Y \to X$, let us look at the definition of the map $H^{n-p}(X, \Omega_X^{n-p}) \to H^{n-p}(Y, \Omega_Y^{n-p})$. Start with the map of differentials $j^* \Omega_X \to \Omega_Y$. Now applying $\bigwedge^{n-p}(.)$, which commutes with pullbacks yields a map $j^* \Omega_X^{n-p} \to \Omega_Y^{n-p}$. Now push it forward and use adjointness of $j_*, j^*$ to obtain a map $\Omega_X^{n-p} \to j_* j^*\Omega_Y^{n-p} \to j_* \Omega_Y^{n-p}$. Now applying $H(X, .)$ gives the required map.

Now in the given problem, where $Y = \left\{P \right \}$, $p= n$, the above map is $H^0(X, \mathcal O_X) \to H^0(P, O_P)$, which is actually the identity map on $k$. Let $\overline{x} \in H^0(X, \mathcal O_X)'$ be the element we get by composing with $t_P$. Then $\overline x(1) = t_P(1)$. Identify $H^0(X, \mathcal O_X)$ with $\text {Hom}(\omega_X, \omega_X)$ in the obvious manner. Denote by $x \in \text {Hom}(\omega_X, \omega_X)'$ the element corresponding to $\overline{x} \in H^0(X, \mathcal O_X)'$. Then $x(Id_{\omega_X}) = t_P(1)$.

Let $\varphi:\text {Hom}(\omega_X, \omega_X)\to H^n(X, \omega_X)'$ be the isomorphism given by the dualizing sheaf, which is given by $\varphi(f) = t_X(H^n(f))$. Then $\varphi '$ composed with the canonical identification $i$ of a vector space and its double dual provides an isomorphism of $\text {Hom}(\omega_X, \omega_X)'$ with $H^n(X, \omega_X)$. Let $\zeta \in H^n(X, \omega_X)$ correspond to $x \in \text {Hom}(\omega_X, \omega_X)'$. Then $x = i(\zeta) \circ \varphi$. So $t_P(1)= x(Id_{\omega_X}) = i(\zeta)(t_X) = t_X(\zeta)$, which is what we are looking for.

Note that after we have fixed an isomorphism of $\left\{P \right \}$ with $\text{Spec} \, k$, ($k^{\sim}, t)$ works as a dualizing sheaf for any isomorphism $t: k \to k$ as $k$ vector spaces. So it is possible for $t_P(1)$ to take any value in $k^*$. Now to get the required answer you have to take $t_P(1) = 1$.

• Thanks very much for your answer. I'm afraid I still have a couple of points of confusion though. Firstly, in the third paragraph you mention how the map $H^0(X,\mathcal{O}_X) \to H^0(P,\mathcal{O}_P)$ is the identity on $k$ after fixing isomorphisms. But this is true for any non-zero map between one-dimensional vector spaces. If we are expected to do this, then what is the content of the problem? Also, in the fourth paragraph you refer to the "canonical identification $i$ of a vector space and its dual" - what is that? – user341315 May 22 '16 at 12:53
• That can't exist because a vector space is not canonically isomorphic to its dual - again, there is a choice involved. Finally, in the last paragraph, you make reference to $k^{\sim}$ and $k^*$ - what is the difference between these? – user341315 May 22 '16 at 12:54
• i am sorry, I meant the double dual. $i : V \to V''$ is defined as $i(x) (f) = f(x)$. This is an injective map and hence an isomorphism in the finite dimensional case. – Shubhodip Mondal May 22 '16 at 12:59
• About your first point, I just meant that identify $H^0(X, \mathcal O _X)$ as $k$, which is actually an equality. So I think I should rephrase my wordings. – Shubhodip Mondal May 22 '16 at 13:01
• and about your last point, by $k^*$ i meant the group of units in $k$ – Shubhodip Mondal May 22 '16 at 13:02