Is every multilinear map expressible as a tensor in the tensor product space? I am curious:
given a vector space $V$ with underlying scalar field $K$, and given a multilinear map $f: V \times V \times \cdots \times V \rightarrow K$, is $f$ always expressible as a tensor in the space given by $V^* ⊗ V^* ⊗ \cdots ⊗ V^*$?
For motivation, I am trying to better understand Riemannian metrics as tensors.
Thanks,
 A: Assuming $V$ is finite-dimensional, then yes.
By the universal property of the tensor product, $f$ induces a functional $V^{\otimes n} \to K$, so it induces an element of $(V^{\otimes n})^*$.
We now claim $(V^{\otimes n})^* \simeq (V^*)^{\otimes n}$, even in the more general case that $V$ is a finite-dimensional vector bundle on some manifold (which according to your comment will be useful in the Riemannian metric case).  To see this, we can define a map $(V^*)^{\otimes n} \to (V^{\otimes n})^*$ by
$$\phi_1 \otimes \cdots \otimes \phi_n \mapsto (x_1 \otimes \cdots \otimes x_n \mapsto \phi_1(x_1) \cdots \phi_n(x_n)).$$
(Here the fact that this description is coordinate-free is essential to the proof when $V$ is a vector bundle.)  To see this map is an isomorphism, it suffices to prove it locally on a set where $V$ is trivializable so we can choose a basis $e_1, \ldots, e_m$ of $V$.  Let $\hat e_1, \ldots, \hat e_m$ be the dual basis of $V^*$; then you just need to observe that $\hat e_{i_1} \otimes \cdots \otimes \hat e_{i_n}$ is sent to the functional which sends $e_{j_1} \otimes \cdots \otimes e_{j_n}$ to $\delta_{i_1,j_1} \cdots \delta_{i_n,j_n} = \delta_{(i_1, \ldots, i_n), (j_1, \ldots, j_n)}$.  Thus the image of the basis $\{ \hat e_{i_1} \otimes \cdots \hat e_{i_n} \}$ of $(V^*)^{\otimes n}$ is exactly the dual basis to the basis $\{ e_{j_1} \otimes \cdots \otimes e_{j_n} \}$ of $V^{\otimes n}$.
A: Note that a multilinear functional $\phi\colon V^n\to F$ corresponds precisely to an element of $(V^{\otimes n})^*$.  There is a linear map
$$
\Psi\colon(V^*)^{\otimes n}\to(V^{\otimes n})^*
$$
given by
$$
\Psi\left(\sum_{i=1}^mf_i^1\otimes\cdots\otimes f_i^n\right)(v_1,\dots, v_n) = \sum_{i=1}^m f_i^1(v_1)\cdot\;\cdots\;\cdot f_i^n(v_n)
$$
(which is multilinear).  This map is always an injection, and the two spaces have the same dimension, so if $V$ is finite dimensional then $\Psi$ is an isomorphism.  
If $V$ is infinite dimensional, then $\Psi$ may not be an isomorphism (in fact, it is never an isomorphism).  I will give an example.  Let $\ell_2$ be the space of infinite sequences of real numbers
$$
x = (x_1,x_2,\cdots)
$$
such that
$$
\sum_{i=1}^\infty x_i^2<\infty
$$
We can define a bilinear map $\phi\colon \ell_2\times\ell_2\to\mathbb R$ by
$$
\phi(x, y) = \sum_{i=1}^\infty x_iy_i
$$
I claim that $\phi$ is not $\Psi(t)$ for any $t\in \ell_2^*\otimes \ell_2^*$.  Indeed, suppose that
$$
t = \sum_{i=1}^m f_i\otimes g_i
$$
If $\phi=\Psi(t)$, it means that for all $x,y\in\ell_2$ we have:
$$
\sum_{i=1}^mf_i(x)g_i(y)=\sum_{i=1}^\infty x_iy_i
$$
Now let $e_i\in\ell_2$ be the sequence that has a $1$ in position $i$ and a $0$ everywhere else.  From the formula above, we have that:
$$
\sum_{i=1}^mf_i(e_p)g_i(e_q)=\delta_{pq}
$$
for all $p,q$.  Written in matrix form, we have:
$$
\begin{pmatrix}
f_1(e_1) & \cdots & f_m(e_1) \\
\vdots & \ddots & \vdots \\
f_1(e_N) & \cdots & f_m(e_N)
\end{pmatrix}
\begin{pmatrix}
g_1(e_1) & \cdots & g_1(e_N) \\
\vdots & \ddots & \vdots \\
g_m(e_1) & \cdots & g_m(e_N)
\end{pmatrix}
=
I_N
$$
where $N$ is any positive integer and $I_N$ is the $N\times N$ identity matrix.  Now $I_N$ has rank $N$.  But by the rank-nullity theorem, the rank of the second matrix is at most $m$.  Taking $N>m$, we get a contradiction.  
