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I'm trying to integrate a function over two vectors which lie on the surface of the unit sphere in D dimensions. The function depends only on the difference between the two vectors, and their dot product. Can I make a change of variables to simplify this integral? Thanks for your help.

\begin{equation} \int \int d^D \mathbf{u} \ d^D \mathbf{v} \ f(\mathbf{u}-\mathbf{v}, \mathbf{u} \cdot \mathbf{v}) \delta(\|\mathbf{u}\|^2-1)\delta(\|\mathbf{v}\|^2-1) \end{equation}

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  • $\begingroup$ From the form of your integral, it appears $f$ depends on the difference $\mathbf{u}-\mathbf{v}$, and not just their distance $|\mathbf{u} - \mathbf{v}|$, can you clarify? $\endgroup$ – Willie Wong Nov 18 '19 at 15:02
  • $\begingroup$ Yes, I'm sorry, the integral actually depends on the signed difference. I will update the post to clarify. Thanks @WillieWong $\endgroup$ – user148767 Nov 18 '19 at 16:47
  • $\begingroup$ As @Josiah mentioned in his answer, you have that $\mathbf{u}\cdot\mathbf{v} = 1 - \frac12 (\mathbf{u}-\mathbf{v})\cdot(\mathbf{u}-\mathbf{v})$ so the two arguments of your function $f$ are not independent. Next, given $\mathbf{u}-\mathbf{v}$ with length < 2, and a unit vector $\omega \perp (\mathbf{u}-\mathbf{v}$, you can solve for $$ \mathbf{u} = \sqrt{1 - |\mathbf{u}-\mathbf{v}|^2 / 4} \omega + \frac12 (\mathbf{u}-\mathbf{v}) $$ and similarly $\mathbf{v}$. So you can reparametrize the the product of the two unit spheres (a $2(D-1)$ dimensional manifold) by the product of the... $\endgroup$ – Willie Wong Nov 18 '19 at 17:27
  • $\begingroup$ .. $D$ dimensional ball $(\mathbf{u}-\mathbf{v})$ against the $D-2$ dimensional sphere ($\omega$). The integral against the latter vanishes and just give you its surface area (with the appropriate radius). The integral against the former factor has to be evaluated, with a carefully computed Jacobian. In the generality that you are seeking that's the best that one can do. $\endgroup$ – Willie Wong Nov 18 '19 at 17:29
  • $\begingroup$ This is just what I'm looking for. Thanks again @WillieWong $\endgroup$ – user148767 Nov 18 '19 at 17:59
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Edit: The below answer addresses a separate question than that in the updated question. @WillieWong's comments above are most relevant to the question when the function depends on the difference $\mathbf{u}-\mathbf{v}$ instead of the distance.

'The function depends only on the distance between the two vectors, and their dot product.'

In other words the function can be taken to only be a function of the dot product, or distance (since $\|x-y\|^2=2-2\langle x,y \rangle$, for $x,y$ on the sphere).

One can expand the function $f$ into its Gegenbauer expansion $f(\langle x,y \rangle)=\sum\limits_{k=0}^{\infty} a_k C_{k}(\langle x,y \rangle)$ where $C_{k}(t)$ are Gegenbauer polynomials. Assuming the convergence is good enough to allow exchanging the sum and integral, the addition formula gives $$\int_{\mathbb{S}^{d-1}}\int_{\mathbb{S}^{d-1}} \sum\limits_{k=0}^{\infty} a_k C_{k}(\langle x,y \rangle) dx dy=\sum\limits_{k=0}^{\infty}\sum\limits_{m=1}^{\dim V_k}a_k b_k \left(\int_{\mathbb{S}^{d-1}} Y_{m,k}(x)dx \right)^2,$$

where spherical harmonics are denoted $Y_{m,k}$, $V_k$ is the finite-dimensional space of spherical harmonics of degree $k$, and $b_k$ are constants (which depend on how one normalizes the spherical harmonics $Y_{m,k}$). This reduces the double integral to a sum of integrals. It is not clear if $dx$ denotes the surface measure on the sphere. If so, then all terms with $k\geq 1$ are zero, meaning the integral is just the constant term in the Gegenbauer expansion of $f$.

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    $\begingroup$ Isn't "computing the constant term of the Gegenbauer expansion" basically just computing the integral asked for? $\endgroup$ – Willie Wong Nov 18 '19 at 15:01
  • $\begingroup$ Of course, if the measure is surface measure (normalized to be a probability measure). But it is much simpler to expand the function into Gegenbauer polynomials and compute the value of the integral this way (than computing the ugly multi-dimensional integral). Computing the constant term in this case requires a single integral over $[-1,1]$. Either way, it is a simplification as compared to viewing the integral naively. $\endgroup$ – Josiah Park Nov 18 '19 at 15:51
  • $\begingroup$ I think you misunderstand me. My point is that the action "expand the function into Gegenbauer polynomials" contains, as one of its sub-actions, "compute the value of the integral in question"; so we have a chicken and egg problem: if the OP already has a way to quickly/easily compute the coefficients of the Gegenbauer polynomial expansion, presumably the question would not have been asked! $\endgroup$ – Willie Wong Nov 18 '19 at 15:55
  • $\begingroup$ I do not see any other way to answer this question without additional information. The OP does not mention being aware of the orthogonal polynomials and how they relate to integrating over the sphere. If @ostrich responds it will be more clear. $\endgroup$ – Josiah Park Nov 18 '19 at 15:58
  • $\begingroup$ With that I am in total agreement. $\endgroup$ – Willie Wong Nov 18 '19 at 16:22

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