Relationship between a vector space V and its dual space V*

Question

I am studying relativity, and as you know the theory extensively uses the notion of covariant and contravariant component of vectors.

My question is the following. Let say $\vec{x}$ is a vector which belongs to vector space V with basis vectors ${e}_{i}$. We know that $\vec{x}$ can be written as: ${x}^{i} {e}_{i}$. The same vector $\vec{x}$ can be written in terms of the basis vectors ${e}^{i}$ of the dual space V* as: ${x}_{i} {e}^{i}$, why is this true? Isn't ${e}^{i}$ a basis set for the dual space V* and not the vector space V? How can the same vector belong to both the vector space V and its dual V*? How can $\vec{x}$ which belongs to V be written in terms of the basis vectors of another vector space V*?

The Wikipedia article on Covariant transformation says that:

...so the dual space has the same dimension as the linear space itself. It is "almost the same space", except that the elements of the dual space (called dual vectors) transform covariantly and the elements of the tangent vector space transform contravariantly.

What does "almost the same" mean?

Answer 1

Let $V$ be an $n$-dimensional vector space over your favourite field with basis $e_1,\dotsc,e_n$ und $V^*$ its dual space with dual basis $e^1,\dotsc,e^n$.

Strictly speaking the vector spaces $V$ and $V^*$ do not have any vectors in common. But what is meant by “almost the same” is that there is an isomorphism of vector spaces $\varphi\colon V\rightarrow V^*$ defined by $\varphi(e_i) =e^i$ for $i=1,\dotsc,n$ (and extending by linearity). Under this isomorphism $x = \sum_{i=1}^n x^ie_i$ is mapped to $\varphi(x) = \sum_{i=1}^nx^i\varphi(e_i) = \sum_{i=1}^n x^ie^i$. So in your notation you have $x_i = x^i$ for the coefficients $x_i$ of $\varphi(x)$ with respect to the basis $e^1,\dotsc,e^n$.

Under the isomorphism $\varphi$ you identify $x$ and $\varphi(x)$ with each other, although they are not the same, as they live in different vector spaces.

Beware that the isomorphism $\varphi\colon V\rightarrow V^*$ depends heavily on your chosen basis $e_1,\dotsc,e_n$ of $V$. This means that without a basis of $V$ you cannot (easily) identify $V$ with its dual space $V^*$.

Answer 2

There are three important observations to be made:

1) An isomorphism between $V$ and $V^*$ exists only if the space $V$ is finite dimensional. When the dimension of $V$ is infinite, no such isomorphism exists.

2) As already noted, the isomorphism between $V$ and $V^*$ strongly depends on the basis chosen. If the vector $v\in V$ corresponds to the linear form $\lambda\in V^*$ under some choice of basis, this is no longer so under some other choice. This is very relevant because in some situations the vector spaces in question offer no natural choice of basis, a typical example is that when $V$ is actually defined as some generic subspace of a bigger vector space.

3) If, on the other hand, the vector space $V$ is endowed with some inner product (i.e. a "generalized dot product") $$ \langle\cdot,\cdot\rangle:V\times V\longrightarrow\Bbb R, $$ where in the general theory of vector spaces over a field $F$ the real field $\Bbb R$ is replaced with $F$, then there is a canonical isomorphism between $V$ and its dual, namely $$ v\mapsto \lambda_v,\qquad\text{where $\lambda_v(w)=\langle v,w\rangle$}. $$ As the term "canonical" implies; this isomorphism does not depend on the choice of basis. It's only in this setting that we can truly regard a vector $v\in V$ as a linear form on $V$.

Answer 3

Your $x_i e^i$ is not a vector, but a linear 1-form. The dual space is exatly the space of such linear forms. You can find a good geometrical introduction to this fact and to all the machinery of covariant and contravariant components in the magnificent book : Gravitation, of Misner-Thorne-Wheeler. ( that you can also search in the web).