Decompose the Laplacian

Question

Is there a way to write the negative Laplacian on the 2-sphere as a decomposition of an operator $A$ and its adjoint $A^*$? I am interested in finding such a decomposition, but I could not get one by simple algebraic manipulations. Does anybody know if something like that exists?

So again, I want to get: $- \Delta = A^* A$ on the $\mathbb{S}^2$. In case that you know a way to get this, I would be also interested in possible domains(if this is non-trivial), such that $A$ is closed.

I think, in principle, the chances that something like that exists are fairly good, as the Laplacian is positive and self-adjoint.

Edit: I know that there is a (i.e. unique positive) square root for these kinds of operators, but I am rather looking for an analogue to the decomposition $-\frac{d^2}{dx^2}= (- \frac{d}{dx}) ( \frac{d}{dx}).$ So, I am rather looking for a casual root( in the sense that the root is comfortable for calculations). I hope I could give you an idea, what I am looking for.

Answer 1

Let $B$ be the non-negative square root of the usual self-adjoint realization of $-\Delta$. Then I claim that $A^*A=-\Delta$ precisely if $A=UB$ for a $U$ satisfying $U^*U=1$. In other words, $U$ varies over the isometries $U: L^2\to R(U)$.

Clearly, such an $A$ satisfies $A^*A=-\Delta$. Conversely, if $A^*A=-\Delta$, then $A1=0$ also (I write $1$ for the constant function $f=1$). This follows because $N(A^*)=R(A)^{\perp}$ and $-\Delta 1=0$. On the orthogonal complement $H$ of $1$, I can invert $B$. Let $C$ be this inverse on $H$, and $C1=0$, say. Let $V=AC$. Then $V^*V = CB^2C=P$, the projection onto $H$. So $V$ is a partial isometry with initial space $H$, and $VB=ACB=AP=A$, since $A$ annihilates $H^{\perp}=L(1)$. Since $B1=0$ also, I can replace $V$ by an isometry $U$ here, without destroying the identity $UB=A$.

Answer 2

Let me follow your notations with $\Delta=\sum_{1\le j\le 3}\partial_{x_j}^2$. You have with $r=\Vert x\Vert$ (the Euclidean norm) $$ r^2\Delta=(r\partial_r)^2+r\partial_r+\Delta_{\mathbb S^2},\quad\text{where $\Delta_{\mathbb S^2}$ is the Laplace operator on the sphere.} $$ Considering an homogeneous harmonic polynomial $p_k$ of degree $k$ on $\mathbb R^3$, we have $$ 0=r^2\Delta p_k=(k^2+k)p_k+\Delta_{\mathbb S^2}p_k, $$ and from that identity, it is easy to get that the spectrum of $-\Delta_{\mathbb S^2}$ is $$\{k^2+k\}_{k\in \mathbb N}=\{(k+\frac{1}{2})^2\}_{k\in \mathbb N}-\frac{1}{4}.$$ We have thus $$ -\Delta_{\mathbb S^2}+\frac{1}{4}=\sum_{k\in \mathbb N}(k+\frac{1}{2})^2\mathbb P_k, $$ where $\mathbb P_k$ is the $L^2(\mathbb S^2)$ orthogonal projection on the space $\Sigma_k$ of spherical harmonics with degree $k$ ($\Sigma_k$=restriction to the sphere of harmonic polynomials of degree $k$). As a result, we get $$ -\Delta_{\mathbb S^2}+\frac{1}{4}=\Lambda ^2,\quad \Lambda=\sum_{k\in \mathbb N}(k+\frac{1}{2})\mathbb P_k. $$ Note that the same procedure could be done in any dimension, that $\Lambda$ is an elliptic pseudodifferential operator of order $1$ on the sphere, which is differential only in the case $\mathbb S^1$ with $\Delta_{\mathbb S^1}=d^2/(d\theta)^2$.

Answer 3

Laplace operator on functions (as a positive operator convention) is defined by $\Delta=d^*d$ where $d:C^\infty(M)\to C^\infty(T^*M)$ is the exterior derivative.