Given two Riemannian manifolds $M$ and $N$, of dimension $m$ and $n$ respectively, the product manifold $M\times N$ has a Riemannian structure, and there is a Laplacian operator $\Delta_{M\times N}$ on $\Lambda^k(M\times N)\cong \bigoplus_j \Lambda^jM\otimes \Lambda^{k-j}N.$ We expect the Laplacian to obey a product rule identity $\Delta_{M\times N}\omega\wedge\eta=(\Delta_M\omega)\wedge\eta + \omega\wedge(\Delta_N\eta),$ where $\omega\in\Lambda^jM$ and $\eta\in\Lambda^{k-j}N.$
We've been trying to prove this identity using the identities for the exterior derivative and Hodge star operator on products: $d(\omega\wedge\eta)=d\omega\wedge\eta + (-1)^j\omega\wedge d\eta$ and $*(\omega\wedge\eta)=(-1)^{(m-j)(k-j)}*\omega\wedge*\eta$.
So far we've been unable to get the signs to come out right, terms that should cancel aren't, and terms that should combine don't. We're hoping another set of eyeballs will find the sign error. Or perhaps the formula itself is wrong, or one of our assumptions is wrong?
Instead of posting the general proof, which is very lengthy and nearly illegible, I thought I would post instead a computation of a special case in $\mathbb{R}^3$. Let $f(x)g(y,z)\,dy\in\Lambda^1\mathbb{R}^3\cong \Lambda^0\mathbb{R}\otimes\Lambda^1\mathbb{R}^2.$
- $\Delta(fg\,dy)=(*d*d)(fg\,dy) + (d*d*)(fg\,dy)=(*d*)(g\partial_xf\,dx\wedge dy +f\partial_zg\,dz\wedge dy) + (d*d)(fg\,dz\wedge dx)
=(*d)(g\partial_xf\,dz-f\partial_zg\,dx) + (d*)(f\partial_yg\,dy\wedge dz\wedge dx)=*(g\partial^2_xf\,dx\wedge dz + \partial_xf \partial_yg\,dy\wedge dz-f\partial_y\partial_zg\,dy\wedge dx-f\partial^2_zg\,dz\wedge dx) + d(f\partial_yg)=-g\partial^2_xf\,dy + \partial_xf \partial_yg\,dx+f\partial_y\partial_zg\,dz-f\partial^2_zg\,dy+\partial_xf\partial_yg\,dx+f\partial^2_yg\,dy+f\partial_z\partial_yg\,dz$
Now the Laplacian of a 1-form on $\mathbb{R}^2$ is given by $\Delta(g\,dy)=(\partial^2_y+\partial^2_z)g\,dy$ and the Laplacian on 0-forms on $\mathbb{R}$ is just $\Delta(f)=\partial^2_xf$. So the result I'm expecting is $\Delta(fg\,dy)=(\Delta f)g\,dy+f(\Delta g\,dy)=g\partial^2_xf\,dy+f(\partial^2_y+\partial^2_z)g\,dy.$ If I could flip the sign of the first and fourth terms in 1, and also flip either the second or fifth so they cancel, and also flip either the third or seventh so they cancel, then the result would be proved.
However we can find no sign error which would justify doing so. What we have instead is $\Delta(fg\,dy)=-(\partial^2_xf)g+f(\partial^2_y-\partial^2_z)g\,dy+(\partial_xf\partial_yg\,dx+\partial_xf\partial_yg\,dx)+(f\partial_y\partial_zg\,dz+f\partial_z\partial_yg\,dz)$ How can I make this calculation work out?
Maybe some of my Hodge stars are incorrect? I'm putting $*(dx)=dy\wedge dz,*(dy)=dz\wedge dx,*(dz)=dx\wedge dy,*(dx\wedge dy)=dz,*(dy\wedge dz)=dx,*(dz\wedge dx)=dy,*(dx\wedge dy\wedge dz)=1.$ Those formulas are standard on $\mathbb{R}^3$. But could they somehow be different because I'm decomposing $\mathbb{R}^3$ as a product $\mathbb{R}\times \mathbb{R}^2$, and each product factor has its own orientation?
And though I haven't posted it here due to length, we've also worked out the derivation for arbitrary an $k$-form $\omega\wedge\eta$ on $M\times N$, and I have similar problems there; terms that should combine don't, terms that should cancel don't. Can someone help me find some more minus signs in this calculation?
Best Answer
Edit: As mentioned in the comments, the Laplacian on Riemannian manifolds is defined not by $\Delta=*d*d+d*d*$, but $\Delta=\delta d+d\delta$, where the coderivative $\delta$ is defined by $\delta=(-1)^{nk+n+1}*d*$ for a $k$-form in $n$ dimensions. In 3D, this is the same as $\delta=(-1)^k*d*$.
I want to point out an even simpler case which perhaps highlights where you are going wrong.
The laplacian evaluated on $f(x)$ (Note that in the next few calculations the $(-1)^k$ is symbolic and is acting as an operator, i.e. noncommutative, because I am waiting to see what rank the form is before I evaluate it): $$\Delta f=(-1)^k*d*df + d(-1)^k*d*f=(-1)^k*d*\partial_xf\,dx + d(-1)^k*df\,dx\wedge dy\wedge dz$$ $$=(-1)^k*d\partial_xf\,dy\wedge dz + d(-1)^k*\partial_xf\,dx\wedge dx\wedge dy\wedge dz$$ $$=(-1)^k*\partial^2_xf\,dx\wedge dy\wedge dz=(-1)^0\partial^2_xf=\partial^2_xf$$
Nothing fancy here, just what you'd expect. Now let's try $f(x)\,dy$: $$\Delta f\,dy=(-1)^k*d*df\,dy + d(-1)^k*d*f\,dy$$ $$=(-1)^k*d*\partial_xf\,dx\wedge dy - d(-1)^k*df\,dx\wedge dz$$ $$=(-1)^k*d\partial_xf\,dz + d(-1)^k*\partial_xf\,dx\wedge dx\wedge dz$$ $$=(-1)^k*\partial^2_xf\,dx\wedge dz=-(-1)^1\partial^2_xf\,dy=\partial^2_xf\,dy$$
So the missing minus sign is coming from this mysterious "codifferential" object. Let's try the full calculation now:
$$\Delta(fg\,dy)=((-1)^k*d*d)(fg\,dy) + (d(-1)^k*d*)(fg\,dy)$$ $$=((-1)^k*d*)(g\partial_xf\,dx\wedge dy +f\partial_zg\,dz\wedge dy) + (d(-1)^k*d)(fg\,dz\wedge dx)$$ $$=((-1)^k*d)(g\partial_xf\,dz-f\partial_zg\,dx) + (d(-1)^k*)(f\partial_yg\,dy\wedge dz\wedge dx)$$ $$=(-1)^1*(g\partial^2_xf\,dx\wedge dz + \partial_xf \partial_yg\,dy\wedge dz-f\partial_y\partial_zg\,dy\wedge dx-f\partial^2_zg\,dz\wedge dx) + d(-1)^0(f\partial_yg)$$ $$=g\partial^2_xf\,dy - \partial_xf \partial_yg\,dx-f\partial_y\partial_zg\,dz+f\partial^2_zg\,dy+\partial_xf\partial_yg\,dx+f\partial^2_yg\,dy+f\partial_z\partial_yg\,dz$$ $$=(\partial^2_xf)g\,dy +f(\partial^2_y+\partial^2_z)g\,dy$$
and all is well.