# Can Hilbert spaces generalize non-Euclidean geometry by having the sum of the angles of a triangle not be equal to pi?

I am an amateur mathematician learning new things.

Let A and B be vectors in a Hilbert space. The three vectors A, B and A-B form a triangle. The idea of the angle between two vectors can be captured using the inner product: the arc cosine of the inner product of A and B divided by the product of the norms of A and B is the angle between A and B. Likewise, the angles between A and B-A and B and B-A can be so defined. If the Hilbert space is an n dimensional Euclidean space and the inner product is the dot product, then the sum of the three angles of the triangle will be pi. Is it possible to define an inner product so that the sum of the three angles will not be pi for an n dimensional Euclidean space or for an infinite dimensional function space? Can Hilbert spaces generalize non-Euclidean geometries or just Euclidean geometry?

• Isn't one of the characteristics of the Euclidean space the fact that the measures of angles are translation and scaling invariant? If so, it seems that the inner product being invariant after normalizing and translation will reduce the space to a Euclidean space. I am sure I am ignoring something obvious. What I mean is the sum of the angles of triangles is dependent on the area of the triangle, so doubling the sides would have to change the angle, but for inner products the cosine would be constant, so the angle would be constant. Jul 28 '14 at 22:48
• This isn't enough for an answer, but one suitable generalization might be infinite dimensional manifolds. They have seen some use in e.g. the theory of PDE, namely fluid flow. Jul 29 '14 at 1:05
• @RandomExcess, I’m with you. Isn’t a finite-dimensional subspace of a Hilbert space Euclidean? Jul 29 '14 at 1:55

No. $A$ and $B$ span a $2$ dimensional subspace in $H$, the whole triangle will be in it. Restriction of inner product on $H$ to a $2$-dimensional subspace is very much Euclidean, and so are all triangles in it, with angle sum $180^\circ$.
If one could get non-Euclidean geometry by restricting to a plane in a Hilbert space one could get it just as well by doing it in a $3$-space. Non-Euclidean geometry does not arise from constant positive definite inner products on vector spaces, either finite or infinite dimensional, by restricting to planes. You need either restrictions to non-linear surfaces or "inner products" that depend on a point (Riemannian metrics) to go non-Euclidean.