Can every Bilinear map $B:V\times V \to k$ be written as follows?

Question

I am studying representation theory on my own and encountered tensor product spaces. I learnt that the space of Bilinear maps on $V\times V$, for a vector space $V$ is isomorphic to $V^* \otimes V^*$ with a 'natural' isomorphism given by the map

\begin{equation} \phi: \alpha_1\otimes \alpha_2 \in V^*\otimes V^* \longrightarrow B: B(v,v') = \alpha_1(v)\alpha_2(v') \end{equation}

I am confused if any Bilinear map on $V\times V$ can be written as $\alpha_1(v)\alpha_2(v')$ as it is not obvious to me. Can someone explain this to me? Thanks for your time.

Answer 1

The considered Solution of mine:

The claim that every bilinear map $B: V \times V \rightarrow k$ can be represented as $\alpha_1(v) \alpha_2(v')$ for some $\alpha_1, \alpha_2 \in V^*$ is not generally true.

The isomorphism between the space of bilinear maps and $V^* \otimes V^*$ is based on a specific choice of basis for $V$. Let $\{v_i\}$ be a basis for $V$, and $\{v_i^*\}$ be its dual basis in $V^*$. The isomorphism $\phi: V^* \otimes V^* \rightarrow \text{Bil}(V \times V, k)$ is given by:

$ \phi(\alpha_1 \otimes \alpha_2)(v, v') = \alpha_1(v) \alpha_2(v') $.

However, this representation depends on the choice of basis. Not every bilinear map can be expressed in the form $\alpha_1(v) \alpha_2(v')$ using the same basis. The isomorphism is "natural" in the sense that it relies on the dual space structure, but it doesn't imply a unique representation for every bilinear map.

Answer 2

Here's an example: Let $V = \mathbb{R}^n$ and consider the bilinear map given by the standard inner product. I claim this cannot be written in the form $ \langle v,w \rangle = \alpha_1(v)\alpha_2(w)$.

Why not? Well if it were then any non-zero vector $u$ in the kernel of $\alpha_1$ would satisfy $\langle u,v \rangle = 0$ for all vectors $v$. But this is impossible because the standard inner product is non-degenerate (in particular it's positive definite so $\langle u,u \rangle > 0$).

The bilinear maps $\alpha_1(v)\alpha_2(w)$ span $V^* \otimes W^*$, but typically you need to actually take linear combinations of them. In this example it turns out you need at least $n$ terms of the form $\alpha_1(v)\alpha_2(w)$ in order to write the standard inner product. The number of such terms you need is what's called the "rank" of the bilinear form, and it measures how non-degenerate the form is.

Answer 3

I learnt that the space of Bilinear maps on $V\times V$, for a vector space $V$ is isomorphic to $V^* \otimes V^*$ with a 'natural' isomorphism given by the map \begin{equation} \phi: \alpha_1\otimes \alpha_2 \in V^*\otimes V^* \longrightarrow B: B(v,v') = \alpha_1(v)\alpha_2(v') \end{equation} I am confused if any Bilinear map on $V\times V$ can be written as $\alpha_1(v)\alpha_2(v')$ as it is not obvious to me.

The correct statement is that there is a unique linear function $\phi$ such that $$(\phi(\alpha_1\otimes\alpha_2))(v_1,v_2)=\alpha_1(v_1)\alpha_2(v_2)$$ for all $\alpha_1,\alpha_2\in V^*$ and $v_1,v_2\in V$. You seem to assume that each element of $V^*\otimes V^*$ is of the form $\alpha_1\otimes\alpha_2$, but this is not correct. The truth is that each tensor can be written as the sum of such simple tensors.