# Definiton of Vector

1. Jul 26, 2010

### schwarzschild

In Schutz's treatment of general relativity he defines a one-form as a function which maps a vector to a real number, and then later defines a vector as a linear function that maps one-forms into the reals. So the definitions seem to be circular - is there another way we can define a vector?

2. Jul 27, 2010

### Fredrik

Staff Emeritus
If V is a finite-dimensional vector space over R (the real numbers), it's dual space V* is defined as the set of linear functions from V into R. Since V* is also a finite-dimensional vector space, we can use the same definition to construct its dual space V**. What you're describing is the definition of V**, not V, so there's nothing circular about it.

Note that V is isomorphic to V**. Just define f:V→V** by f(v)v*=v*v for all v* in V*. This f is an isomorphism, and it's the reason why you can think of V** as "the same thing" as V.

In general relativity, V is the tangent space of spacetime M at some point p in M. So there's a different V for each p. In SR, we have the option to instead take spacetime M to be a vector space, and then there's no need to talk about tangent spaces.

One way to define the tangent space: Let C be the set of smooth functions from M into R. Define V to be the set of linear functions v:C→R such that v(fg)=v(f)g(p)+f(p)v(g) for all f,g in C. Define a vector space structure on V by (u+v)(f)=uf+vf and (av)f=a(vf). Each coordinate system defines a basis for this vector space. The basis vectors are the partial derivative operators defined this way:

(Edit: I should have said smooth functions.)

The proof of that involves a trick that you can find in Wald's GR book or Isham's differential geometry book, if you're interested.

Another option is discussed here. The vector spaces defined by these two definitions are isomorphic, so it doesn't matter which one of them we think of as "the" tangent space at p.

One more detail that may be of interest to you:
It actually doesn't have to be an inner product. Any symmetric non-degenerate bilinear form (like the metric tensor) will do.

Last edited: Jul 27, 2010