Laplace-de Rham operator

Question

Consider an operator $\partial = (-1)^k \star^{-1}\,d\star: \Omega^k(\mathbb{R}^n) \to \Omega^{k-1}(\mathbb{R}^n)$. Note that we equivalently can write $\partial = (-1)^{nk + n + 1} \star\,d\star$. Denote $\triangle := (\partial + d)^2$ (this is Laplace-de Rham operator).

a) Verify that $\triangle$ is an operator $\Omega^k(\mathbb{R}^n) \to \Omega^k(\mathbb{R}^n)$.

b) Compute $\triangle(f\alpha)$ for a function $f: \mathbb{R}^n \to \mathbb{R}$ and a differential $1$-form $\alpha$.

c) Find explicit formulas for $\triangle\alpha$ for the cases when $n = 3$ and $k = 0, 1, 2, 3$.

d) For $n=2$ find explicit formulas for $\triangle \alpha$ in polar coordinates for $k = 0, 1, 2$.

Progress so far: I know how to do a). We have $\nabla = \partial^2 + \partial d + d\partial + d^2$. Now$$\partial^2 = \star^{-1}d\star\star^{-1}\,d\star = \star^{-1}d\,d\star = 0$$so $\triangle = \partial d + d\partial$ and this is a map $\Omega^k(\mathbb{R}^n) \to \Omega^k(\mathbb{R}^n)$. But I am stuck on the other ones. Any help would be appreciated.

Answer 1

b.

We compute $\nabla(\alpha)$ for an arbitrary $1$-form $\alpha(x) = \sum_{i=1}^n \alpha_i(x)\,dx_i$. We have$$\partial \hspace{.5mm} d\alpha = \partial\left(\sum_{i \neq j} D_j\alpha_i\,dx_j \wedge dx_i\right) = \partial\left(\sum_{i < j} (D_i\alpha_j - D_j\alpha_i)\,dx_i \wedge dx_j\right)$$$$ = (-1)^{3n+1}\star d\left(\sum_{i<j}(D_i\alpha_j - D_j\alpha_i) \star (dx_i \wedge dx_j)\right)$$$$ = (-1)^{n+1} \star \left(\sum_{i < j} \left((D_i^2 \alpha_j - D_iD_j\alpha_i)\,dx_i \wedge \star\,(dx_i \wedge dx_j) + (D_iD_j\alpha_j - D_j^2\alpha_i)\,dx_j \wedge \star(dx_i \wedge dx_j)\right)\right).$$Now $dx_i \wedge \star(dx_i \wedge dx_j) = \pm\star dx_j$. We can determine the sign by noting $(\omega$ is the volume form$)$$$dx_i \wedge dx_j \wedge \star(dx_i \wedge dx_j) = \omega = dx_j \wedge \star dx_j.$$Hence $dx_j \wedge dx_i \wedge \star(dx_i \wedge dx_j) = -\omega$ so $dx_i \wedge \star(dx_i \wedge dx_j) = -\star dx_j$.Similarly we find $dx_j \wedge \star(dx_i \wedge dx_j) = \star\, dx_i$. Plugging this in,$$\partial \hspace{.5mm} d\alpha = (-1)^{n+1} \star \left(\sum_{i < j} \left((D_i^2 \alpha_j - D_iD_j\alpha_i)\,dx_i \wedge \star(dx_i \wedge dx_j) + (D_iD_j\alpha_j - D_j^2\alpha_i)\,dx_j \wedge \star(dx_i \wedge dx_j)\right)\right)$$$$= (-1)^{n+1+n-1} \sum_{i < j} \left((D_iD_j\alpha_i - D_i^2\alpha_j)\hspace{.5mm}dx_j + (D_iD_j\alpha_j - D_j^2\alpha_i)\hspace{.5mm}dx_i\right)$$$$=\sum_{i \neq j} (D_iD_j\alpha_j - D_j^2\alpha_i)\hspace{.5mm}dx_i.$$We also have$$d(\partial\alpha) = (-1)^{2n+1}d \star d\left(\sum_{i=1}^n \alpha_i \star dx_i\right) = -d \star \left(\sum_{i=1}^n D_i\alpha_i\hspace{.5mm}dx_i \wedge \star dx_i\right)$$$$=-d \star \left(\sum_{i=1}^n D_i\alpha_i\omega\right) = -d\left(\sum_{i=1}^n D_i\alpha_i\right) = -\sum_{i,\hspace{.5mm}j = 1}^n D_i D_j \alpha_i\hspace{.5mm}dx_j.$$Hence$$\Delta\alpha = (d\partial + \partial d)\alpha = \sum_{i \neq j} (D_iD_j\alpha_j - D_j^2\alpha_i)\hspace{.5mm}dx_i - \sum_{i,\hspace{.5mm}j}D_iD_j\alpha_j\hspace{.5mm}dx_i = -\sum_{i,\hspace{.5mm}j} D_j^2 \alpha_i\hspace{.5mm}dx_i.$$It follows that$$\Delta(f\alpha) = -\sum_{i,\hspace{.5mm}j} D_iD_j(f\alpha_j)\hspace{.5mm}dx_i = -\sum_{i,\hspace{.5mm}j}(\alpha_jD_iD_jf + D_ifD_j\alpha_j + D_jfD_i\alpha_j + fD_iD_j\alpha_j)\hspace{.5mm}dx_i.$$

c.

Let us agree to write $\nabla^2$ for the second order differential operator $\nabla^2 = \sum_{j=1}^n D_j^2$. Thus our calculation from part $($b$)$ shows that for $1$-form $\alpha = \sum_{i=1}^3 \alpha_i(x)\,dx_i$ we have$$\Delta\alpha = -(\nabla^2 \alpha_1\hspace{.5mm}dx_1 + \nabla^2\alpha_2\hspace{.5mm}dx_2 + \nabla^2\alpha_3\hspace{.5mm}dx_3).$$Now we claim that $\Delta$ commutes with the dual operator $\star$, i.e. $\star\Delta\alpha = \Delta\star\alpha$. Assuming this for the moment, we get by dualizing that for $\alpha$ the $2$-form$$\alpha = \alpha_1\hspace{.5mm}dx_2 \wedge dx_3 + \alpha_2\hspace{.5mm}dx_3 \wedge dx_1 + \alpha_3\hspace{.5mm}dx_1 \wedge dx_2,$$$$\Delta\alpha = -(\nabla^2\alpha_1\hspace{.5mm}dx_2 \wedge dx_3 + \nabla^2\alpha_2\hspace{.5mm}dx_3 \wedge dx_1 + \nabla^2\alpha_3\hspace{.5mm}dx_1 \wedge dx_2).$$For $\alpha$ a zero form, $\partial \alpha = \pm\star d\star\alpha = 0$ since $\star\,\alpha$ is an $n$-form. Hence$$\Delta\alpha = (d\partial + \partial d)\alpha = \partial(d\alpha)$$$$ = (-1)^{2n+1}\star d(D_1\alpha\hspace{.5mm}dx_2 \wedge dx_3 + D_2\alpha\hspace{.5mm}dx_3 \wedge dx_1 + D_3\alpha\hspace{.5mm}dx_1 \wedge dx_2)$$$$=-(D_1^2 \alpha + D_2^2\alpha + D_3^2\alpha) = -\nabla^2\alpha.$$Dualizing, $$\Delta\alpha(x)\hspace{.5mm}dx_1 \wedge dx_2 \wedge dx_3 = -\nabla^2\alpha\hspace{.5mm}dx_1 \wedge dx_2 \wedge dx_3.$$

It remains to show that $\star\,\Delta = \Delta\star$. Indeed, $$\star\,\Delta\alpha = \star(d\partial + \partial d) = -(\star\, d \star^{-1}d\star + d\star d)$$ while $$\Delta \star \alpha = (d\partial + \partial d)\star = -(d \star^{-1}d \star\star + \star^{-1}d\star d \star).$$The two are equal since $\star^{-1} = (-1)^{k(n-k)}\star$, and $\star\star = (-1)^{k(n-k)}\text{Id}$.

d.

We note that ${\partial\over{\partial r}}$ and ${1\over{r}}{\partial\over{\partial\theta}}$ give an orthonormal basis for each tangent plane at every point $x \in \mathbb{R}^2$. The dual forms are $dr$ and $r\,d\theta$, from which we infer that $\star\, dr = r\,d\theta$ and $\star \,d\theta = -{1\over{r}}dr$. Say that $k=0$ and fix a $0$-form $f$. Then $d(\partial f) = 0$ and$$\partial\,df = \partial(D_rf\,dr + D_\theta f\,d\theta) = -\star d\left(D_rfr\,d\theta - D_\theta f{1\over r}\,dr\right)$$$$=-\star\left(D_rf\,dr\wedge d\theta + rD_rD_rf\,dr \wedge d\theta - D_\theta D_\theta f {1\over r}\,dr\wedge dr\right)$$$$= -\left({1\over r}D_rf + D_rD_rf + {1\over{r^2}}D_\theta D_\theta f\right).$$For the case $k=2$, we may write$$\Delta f \,dr \wedge d\theta = \Delta \star {f\over r} = \star\,\Delta{f\over r} = -\star\left({1\over r}D_r(f/r) + D_rD_r(f/r) + {1\over{r^3}}D_\theta D_\theta f\right)$$$$-\left({1\over r}D_r(f/r) + D_rD_r(f/r) + {1\over{r^3}}D_\theta D_\theta f\right)r\,dr\wedge d\theta = -\left(D_rD_rf - {{D_r f}\over{r}} + {f\over{r^2}} + {1\over{r^2}}D_\theta D_\theta f\right)dr \wedge d\theta.$$When $k=1$ write the arbitrary $1$-form $\alpha = a_rdr +a_\theta d\theta$. We have $$\partial\,d(a_rdr) = \partial D_g a_r\,d\theta \wedge dr=\star\,d{{D_\theta a_r}\over{r}} = -\left({{D_\theta^2 a_r}\over{r^2}}dr + rD_rD_\theta\left({{a_r}\over{r}}\right)d\theta\right).$$Also, $$d(\partial a_r dr) = -d \star d(ra_rd\theta) = -d\star (a_r + rD_ra_r)\,dr \wedge d\theta=-d(a_r/r + D_ra_r)$$$$=-\left({{D_\theta a_r}\over{r}}d\theta + D_\theta D_r a_r d\theta + D_r\left({{a_r}\over{r}}\right)dr + D_r^2a_rdr\right).$$Adding the two gives$$\Delta a_rdr = -\left({{D_\theta^2 a_r}\over{r^2}} + D_r\left({{a_r}\over{r}}\right) + D_r^2a_r\right)dr - D_\theta D_ra_rd\theta.$$Similarly, $$\partial\,d(a_\theta d_\theta) = \partial D_\theta a_r d\theta \wedge dr = -\star d{{D_ra_\theta}\over{r}} = -\star\left({{D_r^2 a_\theta}\over{r}} dr - {{D_ra_\theta}\over{r^2}}dr + {{D_rD_\theta a_\theta}\over{r}}d\theta\right)$$$$=-\left(D_r^2a_\theta d\theta - {{D_r a_\theta}\over{r}}d\theta - {{D_\theta D_r a_\theta}\over{r^2}}dr\right).$$Also, $$d(\partial a_\theta d_\theta) = -d \star d(a_\theta/r\,dr) = d \star {{D_\theta a_\theta}\over{r}}dr\wedge d\theta = d{{D_\theta a_\theta}\over{r^2}}$$$$ = {{D_\theta^2 a_\theta}\over{r^2}}d\theta + \left({{D_\theta D_ra_\theta}\over{r^2}} - 2D_\theta{{a_\theta}\over{r^3}}\right)dr.$$Adding the two gives$$\Delta a_\theta d_\theta = 2\left({{D_\theta D_ra_\theta}\over{r^2}}-{{D_\theta a_\theta}\over{r^3}}\right)dr -\left(-{{D_\theta^2 a_\theta}\over{r^2}} - D_r\left({{a_\theta}\over{r}}\right) + D_r^2a_\theta\right)d\theta.$$Adding the two gives the final answer.