I've been reading about probability on manifolds. What bothers me is that there's no clear way to generalize the Gaussian distribution to manifolds. If we instead assign a probability distribution to each tangent space, and make sure these probability distributions vary smoothly, then this gives us a very clear way to generalize probability distributions to manifolds. Does such a concept exist in literature?

I want to emphasize I'm being very loose with my definitions. What does it mean to assign a probability distribution to a manifold? Well, I suppose for each tangent space $T_xM$, we assign to it a probability distribution $p_x :T_xM \to [0,\infty)$ and $p_x$ "varies smoothly" for $x \in M$. Please understand, this is a soft question and I am being loose with my definitions since I am trying to get a better idea of what's going on.

Any suggestions of papers or textbooks, I would greatly appreciate.

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    $\begingroup$ For a Riemannian manifold, one can take a natural Gaussian distribution on each tangent space (since a symmetric bilinear form is exactly the data needed to specify a mean zero Gaussian distribtuion). $\endgroup$
    – Will Sawin
    Mar 5, 2022 at 4:05
  • $\begingroup$ @WillSawin can you elaborate on this? You define a probability distribution on a manifold it seems. Is there a paper that talks about this? $\endgroup$ Mar 5, 2022 at 4:11
  • $\begingroup$ The ordinary Gaussian distribution is a probability distribution on $\mathbb{R}$, not a probability distribution on the tangent spaces of $\mathbb{R}$. The standard way to generalize to manifolds is via integration against a top degree differential form (just the standard volume form in the case of $\mathbb{R}$). If you truly want probability distributions on the fibers of a vector bundle $E \to M$, take a section of the top degree exterior algebra of the bundle of vertical forms. $\endgroup$ Mar 5, 2022 at 4:48
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    $\begingroup$ The heat kernel on Euclidean space is the family of Gaussian probability measures at a given center. S.-T. Yau proved that, on complete Riemannan manifold with Ricci curvature bounded below, the heat kernel consists of probability measures. So this could be what you want. Together with result of M. Gaffney, the heat kernel under the same conditions even defines a Feller process $\endgroup$ Mar 5, 2022 at 9:56
  • $\begingroup$ (I think I mixed up Yau and Gaffney; Gaffney proved the conservation of unit mass) $\endgroup$ Mar 7, 2022 at 8:28

1 Answer 1


First let’s generalize the Gaussian distribution to $\mathbb{R}^n$: we can define the Gaussian centered at $\mu\in \mathbb{R}^n $ with standard deviation $\sigma\in \mathbb{R}$ as the unique distribution which is rotationally symmetric about $\mu$ and has $$P[|x-\mu|<r]=\frac{\int_{u=0}^r e^{-u^2/2\sigma^2}dV(u)}{\int_{u=0}^\infty e^{-u^2/2\sigma^2}dV(u)}$$ where $V(u)$ is the hypervolume of the solid ball of radius $u$ in $R^n$. This is the same as the product of the Gaussian distributions with standard deviation $\sigma$ in each dimension.

So for a suitable Riemannian manifold $M$, let $V_\mu(u)$ be the volume of the ball $B(\mu,u)$ of radius $u$ around center $\mu$. Then we can define the Gaussian centered at $\mu\in M$ with standard deviation $\sigma$ to be the unique distribution with $$P[d(x,\mu)<r]=\frac{\int_{u=0}^r e^{-u^2/2\sigma^2}dV_\mu(u)}{\int_{u=0}^\infty e^{-u^2/2\sigma^2}dV_\mu(u)}$$ and where any regions $S,S’$ of equal volume in $B(\mu,r)-B(\mu,r’)$ have probabilities that differ by a ratio of at most $e^{-(r^2-r’^2)/2\sigma^2}$.

This should apply whenever $V_\mu(r)$ is unbounded and has a constant $C$ with $V_\mu(r)<C e^r$. (The second condition may hold for all Riemannian manifolds.) For that matter, it should also apply to any metric spaces which meet the above criteria and where $V_\mu(r)$ is a differentiable function of $R$. If you also want the distributions to vary smoothly with $\mu$, then you should probably require that $M$ has no conjugate points.

  • $\begingroup$ I believe you might as well just take the pushforward of the Gaussian on the given tangent space by the exponential map, no need for any conditional assumptions. $\endgroup$ Mar 5, 2022 at 9:52
  • $\begingroup$ @QuartoBendir, yes, you can do that, but I reformulated the answer to avoid the tangent space entirely, which I find more satisfying. $\endgroup$
    – user44143
    Mar 5, 2022 at 9:56
  • $\begingroup$ It isn't clear to me that your definition is well defined. Is it? $\endgroup$ Mar 5, 2022 at 9:59
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    $\begingroup$ @SpencerKraisler, I enjoyed writing this in the spirit of Busemann’s approach to differential geometry, which doesn’t require manifold-ness at all. $\endgroup$
    – user44143
    Mar 8, 2022 at 21:29
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    $\begingroup$ Not the way Busemann did it! The Busemann function comes from a large and successful research program which deserves to be better known — check out The Geometry of Geodesics. $\endgroup$
    – user44143
    Mar 8, 2022 at 21:49

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