Wick theorem for uniform measure on n-sphere?

Question

Let $S_n$ be a unit n-sphere in $\mathbb{R}_n$: $$ \sum_{i=1}^n x_i^2 = 1 $$

How to generate a probability measure $\mu$ that is uniform on $S_n$?

Well, there is an answer long time ago: You essentially take $y_i$ to be independent random Gaussian variables (with the same mean and variance); in the end you normalize them $x_i = y_i / \sqrt{\sum_{i=1}^n y_i^2}$ to get $\mu$.

One may speculate that the Wick theorem for Gaussian distribution still holds for $\mu$, i.e. correlations can be factorized into pairs, for example
$$ \mathbb{E}[x_i x_j x_k x_l] = \mathbb{E}[x_i x_j ] \mathbb{E}[x_k x_l ]+ \mathbb{E}[x_i x_k ] \mathbb{E}[x_j x_l ] + \mathbb{E}[x_i x_l ] \mathbb{E}[x_j x_k ] $$

The rationale is that we can first generate the independent normal random variable $y_i$. These $y_i$s satisfy Wick theorem, and we normalize correlators after the factorization.

Is this true? Can we prove that Wick theorem for $\mu$ using the rotational invariance?

Update: According to joriki, the Wick theorem above does not hold. However, using the calculation by ben, we have a modified Wick contraction: $$ \mathbb{E}_{\mathbb{S}^{n-1}}[x_i x_j x_k x_l] = \frac{1}{f(2,n)} \mathbb{E}_{\mathbb{R}^{n}}[x_i x_j x_k x_l]\\ = \frac{1}{f(2,n)} ( \mathbb{E}_{\mathbb{R}^{n}}[x_i x_j ] \mathbb{E}_{\mathbb{R}^{n}}[x_k x_l ]+ \mathbb{E}_{\mathbb{R}^{n}}[x_i x_k ] \mathbb{E}_{\mathbb{R}^{n}}[x_j x_l ] + \mathbb{E}_{\mathbb{R}^{n}}[x_i x_l ] \mathbb{E}_{\mathbb{R}^{n}}[x_j x_k ]) \\ = \frac{f(1,n)^2}{f(2,n)} ( \mathbb{E}_{\mathbb{S}^{n-1}}[x_i x_j ] \mathbb{E}_{\mathbb{S}^{n-1}}[x_k x_l ]+ \mathbb{E}_{\mathbb{S}^{n-1}}[x_i x_k ] \mathbb{E}_{\mathbb{S}^{n-1}}[x_j x_l ] + \mathbb{E}_{\mathbb{S}^{n-1}}[x_i x_l ] \mathbb{E}_{\mathbb{S}^{n-1}}[x_j x_k ]) $$ For a general 2k point correlator, we need a factor of $\frac{f(1,n)^k}{f(k,n)}$. The contraction rules remain the same as the Wick theorem.

Answer 1

You can indeed infer a Wick's theorem for spheres from the usual one for Gaussians.

As we know that only an even number of contractions gives a non-zero result, let us consider a $2k$-point function for the standard Gaussian in $\mathbb R^n$ $$ \begin{align*} \mathbb E_{\mathbb R^n}(x^1_{i_1}\ldots x^{2k}_{i_{2k}}) &= \frac{1}{(2\pi)^{n/2}}\int_{\mathbb R^n} d^n x\, e^{-x\cdot x/2}x^1_{i_1}\ldots x^{2k}_{i_{2k}}\\ & =\frac{1}{(2\pi)^{n/2}}\int_0^\infty dr\, r^{2k+n-1}e^{-r^2/2} \int_{S^{n-1}} d\Omega\, \hat x^1_{i_1}\ldots \hat x^{2k}_{i_{2k}}, \end{align*} $$ where we wrote out the integral in spherical coordinates and split coordinates into a unit vector and a radius as $x_i = r \hat x_i$.

The second integral is almost the expectation value we are after but it needs normalization by the volume of the sphere $V_{S^{n-1}} = \frac{2\pi^{n/2}}{\Gamma(n/2)}$. Thus, we find $$ \mathbb E_{\mathbb R^n}(x^1_{i_1}\ldots x^{2k}_{i_{2k}})=f(k,n)\, \mathbb E_{S^{n-1}}(\hat x^1_{i_1}\ldots \hat x^{2k}_{i_{2k}}),\qquad f(k,n)= \frac{V_{S^{n-1}}}{(2\pi)^{n/2}}\int dr\, r^{2k+n-1}e^{-r^2/2}, $$ which we can simplify as $$ f(k,n) = \frac{2^k\Gamma(k+n/2)}{\Gamma(n/2)} = \frac{(2(k-1)+n)!!}{(n-2)!!}. $$

As a check we can find the 2-point function $$ \mathbb{E}_{S^{n-1}}(x_i x_j) = \frac{1}{f(1,n)}\delta_{ij} = \frac{\delta_{ij}}{n}, $$ which is correct.

Answer 2

Nice idea, but unfortunately not.

If we choose $x_i=x_j=y$ and $x_k=x_l=z$, the second and third terms vanish by symmetry, and we’re left with $\mathbb E\left[y^2z^2\right]=\mathbb E\left[y^2\right]\mathbb E\left[z^2\right]$, which says that $y^2$ and $z^2$ are uncorrelated. That’s indeed the case for the Gaussian distribution, but not for the uniform distribution on the sphere, where these are negatively correlated. This correlation is introduced by the normalization.