Suppose that $X_i$ are i.i.d. N(0,1) random variables, and set $S = \sum_{i=1}^n X_i^2$. Then $S \sim \chi^2_{(n)}$, the $\chi^2$ distribution with $n$ degrees of freedom. Compute the induced probability measure on $\mathbb{R}^n$: $$\mu^s_n(A) := P((X_1, \dots, X_n) \in A | S = s),$$ where $A \subset \mathbb{R}^n$ is Borel-measurable.
First, a proposed solution. Let $X = (X_1, ..., X_n)$. Then $S = \|X\|^2$. Moreover, for any rotation matrix $O$, $OX$ has the same distribution as $X$. Thus, for any rotation matrix $O$, $$\mu^s(OA) = P(X \in OA | S = s) = P(O^{-1}X \in A | S = s) = \mu^s(A),$$ so $\mu$ is rotation invariant. Moreover, given that $S = s$, $X$ must lie on the sphere of radius $\sqrt{s}$, and $\mu^s(\mathbb{S}_{\sqrt{s}}^{n-1}) = 1$, where $\mathbb{S}_{\sqrt{s}}^{n-1}$ denotes the $n-1$-sphere in $\mathbb{R}^n$ of radius $\sqrt{s}$. Thus $\mu$ is a rotation-invariant probability measure which takes full mass on the $n-1$-sphere; by uniqueness of such measures, $\mu$ must be the uniform distribution on the $n-1$-sphere.
The problem I have is computing this explicitly. This measure is singular with respect to $\mathbb{R}^n$-Lebesgue measure, though theorems on regular conditional probabilities tell me this is a well-defined measure for each $s$.
Problem
Take $n = 2$, let $R^2 = X_1^2 + X_2^2$, $\theta = \arctan(X_2/X_1)$. We have $$\left|\begin{pmatrix} \frac{\partial R}{\partial X_1} & \frac{\partial R}{\partial X_2} \\ \frac{\partial \theta}{\partial X_1} & \frac{\partial \theta}{\partial X_2}\end{pmatrix}\right| = \left|\begin{pmatrix} \frac{X_1}{\sqrt{X_1^2+X_2^2}} & \frac{X_2^2}{\sqrt{X_1^2+X_2^2}} \\ \frac{-X_2}{X_1^2+X_2^2} & \frac{X_1}{X_1^2+X_2^2}\end{pmatrix}\right| = \frac{1}{R}.$$ Hence, using independence of $X_1$ and $X_2$ and the density for a standard normal, $$f(R,\theta) = f_{X_1,X_2}(X_1, X_2)\left|\frac{\partial (R,\theta)}{\partial (X_1,X_2)}\right|^{-1} = \frac{1}{2\pi}e^{-R^2}R.$$ From this we see further that $R$ and $\theta$ are actually independent, with $\theta$ being uniform on $(0,2\pi)$ (I know this is not the case in higher dimensions). Thus, the desired density should be $$f(R,\theta | S = s) = \frac{f(s,\theta)}{f(s)} = \frac{1}{2\pi},$$ but this doesn't take into account the radius of the sphere.
I suspect the problem is with having a density on an $n-1$ dimensional submanifold and not taking into account the volume form correctly, but I'm not sure how to fix this. I'd also like some rigorous explanation of why (or if!) I am allowed to just apply the rules for conditional density like I did. Any help appreciated.
Actually, when $n=2$, the density of $(R,\Theta)$ is the function $f_{R,\Theta}$ defined by $$f_{R,\Theta}(r,\theta)=r\mathrm e^{-r^2/2}\mathbf 1_{r\gt0}\,\frac1{2\pi}\mathbf 1_{0\leqslant\theta\lt2\pi}. $$ Thus, for every $r\gt0$, the conditional density of $\Theta$ conditionally on $R=r$ is the function $g_{\Theta\mid R=r}$ defined by $$ g_{\Theta\mid R=r}(\theta)=\frac1{2\pi}\mathbf 1_{0\leqslant\theta\lt2\pi}. $$ Since $g_{\Theta\mid R=r}$ does not depend on $r$, one sees that $R$ and $\Theta$ are independent.
Likewise, in every dimension $n\geqslant2$, the spherical coordinates read $$ (X_1,\ldots,X_n)=R\cdot Z, $$ where $R$ is in $[0,\infty)$ and $Z$ in $S^{d-1}$. Then, $R$ and $Z$ are independent and $Z$ is uniformly distributed on $S^{d-1}$.