Now that we can define the inner product of two vectors using a metric , we want to generalize this to apply to vector fields.
This should be pretty straightforward: if and are vector fields on an open region it gives us vectors and at each . We can hit these pairs with to get , which is a real number. Since we get such a number at each point , this gives us a function .
That this is a bilinear function is clear. In fact we’ve already implied this fact when saying that is a tensor field. But in what sense is it an inner product? It’s symmetric, since each is, and positive definite as well. To be more explicit: with equality if and only if is the zero vector in . Thus the function always takes on nonneative values, is zero exactly where is, and is the zero function if and only if is the zero vector field.
What about nondegeneracy? This is a little trickier. Given a nonzero vector field, we can find some point where is nonzero, and we know that there is some such that . In fact, we can find some region around where is everywhere nonzero, and for each point we can find a such that . The question is: can we do this in such a way that is a smooth vector field?
The trick is to pick some coordinate map on , shrinking the region if necessary. Then there must be some such that
because otherwise would be degenerate. Now we get a smooth function near :
which is nonzero at , and so must be nonzero in some neighborhood of . Letting be this coordinate vector field gives us a vector field that when paired with using gives a smooth function that is not identically zero. Thus is also nonzero, and is worthy of the title “inner product” on the module of vector fields over the ring of smooth functions .
Notice that we haven’t used the fact that the are positive-definite except in the proof that is, which means that if is merely pseudo-Riemannian then is still symmetric and nondegenerate, so it’s still sort of like an inner product, like an symmetric, nondegenerate, but indefinite form is still sort of like an inner product.