MHB Why was there no solution to this week's problem?

[Chris L T521]

Here is this week's problem!

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Problem: Let $(M,g)$ be an oriented Riemannian manifold with the standard Euclidean metric, and let $\omega\in\mathcal{A}^k(M)$ be a $k$ form. We define the Laplace-Beltrami operator as a map $\Delta:\mathcal{A}^k(M)\rightarrow\mathcal{A}^k(M)$ defined by $\Delta = dd^{\ast}+d^{\ast}d$, where
\[d^{\ast}\omega = (-1)^{n(k+1)+1}\ast d\ast \omega\]
with $\ast:\bigwedge^k T^{\ast}M\rightarrow\bigwedge^{n-k} T^{\ast}M$ denoting the Hodge star operator. When $k=0$, show that $\Delta$ agrees with the typical Laplacian on real valued functions: $\displaystyle\Delta u = -\text{div}\,(\text{grad}\,u)= -\sum_{i=1}^n\frac{\partial^2u}{(\partial x^i)^2}$.

Remark: Note for $k=0$, $\mathcal{A}^0(M)=C^{\infty}(M)$, the set of continuous infinitely differentiable functions on the manifold $M$.

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Here's a hint/suggestion.

Spoiler

Use the fact that if $f\in \mathcal{A}^0(M)=C^{\infty}(M)$, then $d^{\ast} f = 0$. While doing this problem, assume WLOG that $\dim M = n$.

 

[Chris L T521]

No one answered this week's problem. :(

Here's the solution.

Spoiler

Proof: Let $u\in\mathcal{A}^0(M)=C^{\infty}(M)$, and suppose WLOG that $\dim M = n$. We observe that\[\Delta u = dd^*u+d^*du = d^*du\]
since $d^*u=0$ for $u\in\mathcal{A}^0(M)$. In the case of $g$ being the standard Euclidean metric, we see that
\[\begin{aligned}d^*du &= d^*\left(\sum_{i=1}^n\frac{\partial u}{\partial x^i}dx^i\right)\\ &= (-1)^{2n+1}\ast d\ast\left(\sum_{i=1}^n\frac{\partial u}{\partial x^i}dx^i\right)\\&= -\ast d\left(\frac{\partial u}{\partial x^1}dx^2\wedge\cdots\wedge dx^n+\cdots+(-1)^{n+1}\frac{\partial u}{\partial x^n}dx^1\wedge\cdots\wedge dx^{n-1}\right)\\ &= -\ast\left(\sum_{i=1}^n\frac{\partial^2 u}{\partial x^i\partial x^1}dx^i\wedge dx^2\wedge\cdots\wedge dx^n+\cdots+(-1)^{n+1}\sum_{i=1}^n\frac{\partial^2 u}{\partial x^i\partial x^n}dx^i\wedge dx^1\wedge\cdots\wedge dx^{n-1}\right)\\&= -\ast\left(\sum_{i=1}^n\frac{\partial^2u}{(\partial x^i)^2}dx^1\wedge\cdots\wedge dx^n\right)\\ &= -\sum_{i=1}^n\frac{\partial^2u}{(\partial x^i)^2}\\ &= -\text{div}\,(\text{grad}\,u).\end{aligned}\]
This completes the proof. QEDRemark: In computing $\ast dx^i$ as seen above, the $n-1$-forms were ordered in the following fashion:
\[\begin{aligned}\ast dx^1 &= dx^2\wedge dx^3\wedge dx^4\wedge dx^5\wedge\cdots\wedge dx^n\\ \ast dx^2 &= -dx^1\wedge dx^3\wedge dx^4\wedge dx^5\wedge\cdots\wedge dx^n\\ \ast dx^3 &= dx^1\wedge dx^2\wedge dx^4\wedge dx^5\wedge\cdots\wedge dx^n\\ \ast dx^4 &= -dx^1\wedge dx^2\wedge dx^3\wedge dx^5\wedge\cdots\wedge dx^n\\ &{\color{white}.}\,\,\,\!\!\vdots\\ \ast dx^{n-1} &= (-1)^ndx^1\wedge dx^2\wedge dx^3\wedge\cdots\wedge dx^{n-2}\wedge dx^n\\ \ast dx^n &= (-1)^{n+1}dx^1\wedge dx^2\wedge dx^3\wedge dx^4\wedge\cdots\wedge dx^{n-1}.\end{aligned}\]
The alternating sign pattern becomes obvious after computing $\ast dx^i$ in $\bigwedge^3T^{\ast}M$, $\bigwedge^4T^{\ast}M$ and $\bigwedge^5T^{\ast}M$.