# Soft Question Hilbert Space Geometry

Just a quick question about the geometry of Hilbert spaces from an intuitive standpoint. Maybe just assuming we're working with $L^2$ would simplify the situation. Basically, in something like $\mathbb{R}^2$ we have the situation that $\cos(\theta)=\frac{\langle a,b\rangle}{\vert a\vert\cdot\vert b\vert}$, and the idea of an angle between vectors is very meaningful, geometrically. We can easily extend this idea to $\mathbb{R}^n$ because when we talk about the angle between two vectors, we mean we are choosing the plane that both of them lie in, and picking the vector in there. But what does this really mean in a Hilbert space like $L^2$? I have a good intuition about functions, and about geometry (topology) separately, but not really the "geometry of functions".

Now, there may be no visualization of this in $L^2$, and I'm not asking for one, but is there any sense to doing geometry (i.e. actual polygons, things like that) in a space like $L^2$? Also, what sort of applications do ideas like this have in functional analysis? Are we ever interested in ideas like "planes" of functions, polygons, surfaces, solids, etc.? What do we really mean by angles, projections, normal vectors? And do these sorts of things ever have any sort of interesting relationships?

I'm primarily asking for an intuitive idea here. It's easy to just do the math, prove theorems about inner products, norms, etc. Maybe geometry gives some clues or intuitive ideas when doing functional analysis?

• "... we mean we are choosing the plane that both of them lie in, and picking the vector in there. But what does this really mean in a Hilbert space like L2?" Why should it mean anything different? – yasmar Dec 1 '10 at 19:40
• Just that there really isn't a geometric idea of a plane in $L^2$, i.e. what do all the functions in a plane have in common? – Jonathan Beardsley Dec 3 '10 at 6:46

## 3 Answers

I think in infinite-dimensional spaces one should interpret the angle as a measure of how correlated two functions are. The intuition manifests itself most clearly when the functions are random variables of mean zero; then their inner product is precisely their covariance, which is zero when they are independent (that is, uncorrelated).

Certainly one of the most important thing about Hilbert spaces is that there is a good notion of projection, and things one would expect to intuitively be true about projection (for example, that the projection of a vector onto a subspace is the part of the subspace closest to the vector) are actually true because it turns out the inner product makes everything work. So in that sense, one can sometimes reliably use finite-dimensional intuition in the infinite-dimensional case.

Perhaps one can also draw intuition from quantum mechanics, where the inner product is a "probability amplitude." I don't know enough about quantum mechanics to really elaborate on this point, though.

• That's a nice interpretation, that is, the probability idea. Then we do in fact have an idea of "closeness." – Jonathan Beardsley Dec 2 '10 at 18:54

One motivation for the $L^2$ inner product is as follows: Imagine sampling your functions f and g at N equally-spaced points, and putting those values into vectors $\vec{f}$ and $\vec{g}$. Then (modulo certain technical assumptions) in the limit as you take more and more samples, the angle between the approximating vectors with a standard dot product approaches the angle between the functions with the $L^2$ inner product. It's not quite rigorous, but a useful mental model is to think of each different x in the domain as representing a different orthogonal direction in your function space, and then the value f(x) is the "coordinate" of f in the x direction.

Let us say we are dealing with real Hilbert space $X=L^2[0,1]$. Any two functions $f,g\in X$ are contained in the plane $P$ spanned by $f,g$. This plane is isometrically isomorphic to the Euclidean space $\mathbb{R}^2$; this means that $P$ is spanned by two orthonormal vectors $e_1, e_2$ (any two orthonormal vectors), and the linear map $T$ that maps $e_1$ to $(1,0)$ and $e_2$ to $(0,1)$ preserves distance and inner product (and hence also angle). The angle between $f$ and $g$ is precisely the angle between $Tf$ and $Tg$ in $\mathbb{R}^2$.

If $f, g$ are linearly independent, $e_1,e_2$ can be obtained from them by Gram-Schmidt process, i.e. $$e_1= \frac{f}{\|f\|}, e_2=\frac{g-(e_1,g)e_1}{\|g-(e_1,g)e_1\|}.$$