The Unapologetic Mathematician

Mathematics for the interested outsider

Inner Products of Vector Fields

Now that we can define the inner product of two vectors using a metric g, we want to generalize this to apply to vector fields.

This should be pretty straightforward: if v and w are vector fields on an open region U it gives us vectors v_p and w_p at each p\in U. We can hit these pairs with g_p to get g_p(v_p,w_p), which is a real number. Since we get such a number at each point p, this gives us a function g(v,w):U\to\mathbb{R}.

That this g is a bilinear function is clear. In fact we’ve already implied this fact when saying that g is a tensor field. But in what sense is it an inner product? It’s symmetric, since each g_p is, and positive definite as well. To be more explicit: g_p(v_p,v_p)\geq0 with equality if and only if v_p is the zero vector in \mathcal{T}_pM. Thus the function g(v,v) always takes on nonneative values, is zero exactly where v is, and is the zero function if and only if v is the zero vector field.

What about nondegeneracy? This is a little trickier. Given a nonzero vector field, we can find some point p where v_p is nonzero, and we know that there is some w_p such that g_p(v_p,w_p)\neq0. In fact, we can find some region U around p where v is everywhere nonzero, and for each point q\in U we can find a w_q such that g_q(v_q,w_q)\neq0. The question is: can we do this in such a way that w_q is a smooth vector field?

The trick is to pick some coordinate map x on U, shrinking the region if necessary. Then there must be some i such that

\displaystyle g_p\left(v_p,\frac{\partial}{\partial x^i}\bigg\vert_p\right)\neq0

because otherwise g_p would be degenerate. Now we get a smooth function near p:

\displaystyle g\left(v,\frac{\partial}{\partial x^i}\right)

which is nonzero at p, and so must be nonzero in some neighborhood of p. Letting w be this coordinate vector field gives us a vector field that when paired with v using g gives a smooth function that is not identically zero. Thus g is also nonzero, and is worthy of the title “inner product” on the module of vector fields \mathfrak{X}(U) over the ring of smooth functions \mathcal{O}(U).

Notice that we haven’t used the fact that the g_p are positive-definite except in the proof that g is, which means that if g is merely pseudo-Riemannian then g 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.

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September 30, 2011 Posted by | Differential Geometry, Geometry | 2 Comments