Definition of dual tensors

Question

It is a clear fact that the dual of vectors is a map taking vectors to real numbers: $V^\ast: \rightarrow$$\mathbb{R}$. Following that how to understand intuitively the notion of the dual tensors? Does it take tensors to real numbers?

Answer 1

It may not be worth it thinking of it that way. Suppose you have an element of $V\otimes V$. This is a tensor. Depending on how you think of tensors, this is an element of a tensor product, a multilinear map from $V^*\times V^* $ to a field, let's stick with $\mathbb{R}$, or a collection of numbers $\{T_{ab}\}$ transforming as a (2,0) covariant object under a change of basis. These different way of thinking about a tensor aren't always equivalent, but for finite dimensional vector spaces and infinite fields they are.

Back to your question: if you understand the dual of $V$, a good way of thinking of the dual of $T\in V\otimes V$ is as an element $T^*$ of $(V \otimes V)^*\simeq V^* \otimes V^*$, which you can identify with a bilinear map from $V \times V$ to $\mathbb{R}$. Similarly the dual of an element in $V^*\otimes V^*$ is an element of $(V^*\otimes V^*)^*\simeq V^{**} \otimes V^{**}\simeq V\otimes V$ which you can think of as a bilinear map from $V^*\times V^*$ to $\mathbb{R}$.

What I am trying to say is that by understanding $V^*$, the isomorphism $V^{**} \simeq V$, so that element of $V$ can be identified with linear maps from $V^* $ to $\mathbb{R}$, and the fact that (up to isomorphisms) taking the dual commutes with with taking the tensor product you have a way of understanding the dual of tensor of arbitrary ranks.

A basis $\{e _a\}$ of $V$ canonically determines a basis $\{e^a\}$ of $V^*$ and in turn bases of any tensor product of $V$ and $V^*$, hence by knowing how to take the dual of elements of $V$ and $V^*$, which is simply the map $ e_a \leftrightarrow e^a$ you can take the dual of an arbitrary tensor.