I am trying to understand the proof of the Holomorphic Lefschetz fixed point formula on page 426 in Griffiths and Harris. However, I find their use of currents extremely confusing. They seem to go back and forth between currents and forms without a thought, which is confusing to me as I don't know much about currents.

Definitions: given a manifold $M$ and a smooth $p$-form $\phi$, the current induced by $\phi$ is the linear functional $T_\phi:\Omega^{n-p}_c(M)\to\mathbb{R}$ given by $T_\phi(\omega)=\int_M\phi\wedge\omega$. Here $\Omega^{n-p}_c(M)$ denotes compactly supported smooth $(n-p)$-forms. A current $T$ restricts to an open subset $U$ by extending a compactly supported form $\omega$ on $U$ by $0$ and then applying $T$. In my case I am interested in complex currents, which are defined analagously with respect to Dolbeault cohomology.

As best as I can tell, the situation is that we have we have a compact manifold $M$ of dimension $n$, a compact submanifold $A$ of dimension $k$, and smooth currents $T_\phi$ and $T_\psi$ where $\phi$ and $\psi$ are closed smooth $k$-forms on $M$. It seems they are implicitly using that if $T_\phi$ restricted to a coordinate neighborhood $U$ of $M$ is zero, then $\int_A\phi=\int_{A-U}\phi$, and further if $T_\phi=T_\psi$ on an open set $W$ such that $A - U \subset W$ then $\int_{A-U}\phi=\int_{A-U}\psi$.

Do the statements above make sense? Are they true?

Or do Griffiths and Harris mean something else? (Note I have used somewhat different notation than they used to not clutter this question. )

BOUNTY: I'm setting a bounty for someone who can give a satisfactory explanation as to why, in the notation of the book,

$$\int_{\Gamma_f}\phi=-\int_{\Gamma_f-\cup B_\epsilon(p_\alpha,p_\alpha)} \bar\partial k$$

This could mean proving or giving a reference to the two statements I stated above, of if those statements are not correct, then proving the above equality using the information given on page 426.

UPDATE: It seems that Poincare duality for noncompact manifolds implies by the hypotheses that on $U$, $\phi=0$ up to an exact form, and on $W$, $\phi=\psi$ up to an exact form. However, $A-U$ will have boundary. So I still haven't solved the problem.


1 Answer 1


First, I will answer your question about the equality you quote from G&H; in the meanwhile, it could happen that I also answer your two general questions about currents.

The setting is the following: we are considering currents in the product space $M\times M$, where $M$ is a $n$-dimensional complex manifold; $\Delta$ is the diagonal, i.e. $$\Delta=\{(x,x)\ :\ x\in M\}$$ and it is obviously isomorphic to $M$ (useless here, however). $T_\Delta$ is the current of integration along the diagonal, so, for every test form $u\in\mathcal{D}^{2n}(M\times M)$ (smooth compactly supported $2n$-forms), $$T_\Delta(u)=\int_\Delta u\;.$$ By employing the decomposition in bidegrees, we can define $T^0_\Delta$ as the component of $T_\Delta$ which acts only on forms of bidegree $(n,n-*)-(0,*)$, i.e. the component of bidimension (bitype in the terminology of G&H) $(0,*)-(n,n-*)$.

Now, $T_\Delta^0$ is a current, but is not represented by integration against a smooth form, as the ones you defined (it is represented by integration against a form with measure coefficients, but let's not enter such details).

By means of the Bochner-Martinelli formula, we can locally define (around fixed points $\{p_\alpha\}$ for $\alpha$ in some set of indexes) currents of compact support (contained in given balls of radius $2\epsilon$ around those points) $k_\alpha=\rho_\alpha k_{BM}$ such that $\overline{\partial}k_\alpha=T_\Delta^0$ on a ball of center $(p_\alpha,p_\alpha)$ and radius $\epsilon$. This means that, for every $u\in\mathcal{D}^{2n}(B_\epsilon(p_\alpha,p_\alpha))$, $$T_{\Delta}^0(u)=\overline{\partial}k(u)$$ (meaning that we take the $\overline{\partial}$ of $k$ as a current and we apply it - which is again a current - to the form $u$).

Now, defining $k=\sum_\alpha k_\alpha$, you obtain a globally defined current on $M\times M$. By inspection, one notices that $k$ is given by integration against a form which is smooth on $M\times M\setminus\Delta$, i.e. there exists $\omega_{BM}\in\Omega^{2n+1}(M\times M\setminus \Delta)$ such that $$k(u)=\int_{M\times M}\omega_{BM}\wedge u$$ (this implies that $\omega_{BM}$ has locally integrable coefficients on $M\times M$, which is true and can be verified by looking at the order of "pole" along $\Delta$).

Therefore, outside $\Delta$, also $\overline{\partial} k$ is represented by integration against some form $\eta\in\Omega^{2n}(M\times M\setminus \Delta)$. The words "outside $\Delta$" are quite important here: we cannot say that there is such an $\eta$ so that $$\overline{\partial}k(v)=\int_{M\times M}\eta\wedge v$$ for every $v\in\mathcal{D}^{2n}(M\times M)$!! We don't know what happens along $\Delta$. We can say, tho $$\overline{\partial}k(v)=\int_{M\times M}\eta\wedge v$$ for every $v\in\mathcal{D}^{2n}(M\times M\setminus\Delta)$, i.e. for forms with compact support outside $\Delta$.

However, a closer inspection of $k_{BM}$, reveals that its $\overline{\partial}$ vanishes outside the diagonal, so the only part which survives is given by $\overline{\partial}\rho_\alpha\wedge k_{BM}$, which is again integrable all over $M\times M$, the "poles" along the diagonal being the same. Therefore, $$v\mapsto\int_{M\times M}\eta\wedge v$$ is a well defined current $T_\eta$ for $v\in\mathcal{D}^{2n}(M\times M)$. But, again, it does not coincide with $\overline{\partial} k$! We have $$\overline{\partial} k= T_\eta+R$$ where $R$ is a current, of which we only know that it is of locally finite mass and supported on the diagonal.

So, summing up, we define the current $$\phi=T_{\Delta}^0-\overline{\partial}k\;.$$ Now, let $u$ be a $2n$-form with compact support in $B_\epsilon(p_\alpha,p_\alpha)$; on such a ball, we arranged things so that $T_{\Delta}^0=\overline{\partial} k=\overline{\partial} k_\alpha$ (the last equality holding only on that ball, not on the whole space). Then we have $$\phi(u)=T_\Delta^0(u)-\overline{\partial}k (u)=0$$ as $u$ has support in $B_\epsilon(p_\alpha,p_\alpha)$.

In terms of $\eta$ and $R$, we see that $$0=T_{\Delta}^0(u)-T_\eta(u)-R(u)$$ The first and last current are supported on $\Delta$, so the contributions outside $\Delta$ all come from $T_\eta$; it means that the support of $T_\eta$ does not intersect $B_\epsilon$. That is $$T_\eta(u)=0$$ for every $u$ a $2n$-form with compact support in $B_\epsilon(p_\alpha,p_\alpha)$. I.e. $$\int_{B_\epsilon}\eta\wedge u=0$$ for every $u$, which easily implies that $\eta\vert_{B_\epsilon}=0$: take $\tau_n$ a function with values in $[0,1]$, which is $1$ inside a ball of radius $\epsilon-\epsilon/n$ and supported in $B_\epsilon$. By dominated convergence $$0=\int_{B_\epsilon}\eta\wedge (\tau_n \eta^*)\to\int_{B_\epsilon}\eta\wedge \eta^*=\|\eta\|_{L^2(B_\epsilon)}^2$$ which then has to be zero, so $\eta\vert_{B_\epsilon}=0$.

Now, $\phi$ is represented by a smooth form (as G&H says) at least on $$U=(M\times M\setminus\Delta)\cup\bigcup_{\alpha} B_{\epsilon}(p_\alpha,p_\alpha)$$ and such a smooth form is our $\eta$: given $u\in\mathcal{D}^{2n}(U)$, we can write $$u=\sum_{\alpha}\sigma_\alpha u + \left(u-\sum_\alpha \sigma_\alpha u\right)=\sum u_\alpha + u'$$ for suitable $\sigma_\alpha$ so that $u_\alpha$ is supported in $B_\epsilon$ and $u'$ is $0$ on (a neighbourhood of) $\Delta$. So $$\phi(u)=T_\Delta^0(u)-T_\eta(u)-R(u)=T_\Delta^0(u')-T_\eta(u')-R(u')=-T_\eta(u')=-T_\eta(u)$$ because $u'$ has support which doesn't meet $\Delta$, where $T_\Delta^0$ and $R$ live.

Again, this implies that the form that represents $\phi$ (which we will denote by $\psi$, so that $\phi=T_\psi$) and the form that represents $-T_\eta$ (i.e. $-\eta$) coincide on $U$.

Now, as $\Gamma_f$ is a $2n$-dimensional submanifold of $U$, we have $$\int_{\Gamma_f}\psi=-\int_{\Gamma_f}\eta$$ but as $\eta$ vanishes on all $B_\epsilon$, this is the same as $$-\int_{\Gamma_f\setminus\bigcup B_\epsilon}\eta\;.$$

Sorry if I changed notation a bit, but I wanted to keep currents and forms as much separated as possible, because you said that it was their interplay that confused you.


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