# Question about tensor products, decomposable tensors, …

I need some help with the following problem:

Let $V_1,\ldots,V_m$ be finite dimensional vector spaces over $\mathbb{K}$.

Let $\varphi \in L(V_1,\ldots,V_m;U)$ such that $Im(\varphi)=U$.

Show that there exists a subspace $K$ of $V_1 \otimes \cdots \otimes V_m$ such that every class of $\frac{V_1 \otimes \cdots \otimes V_m}{K}$ contains a decomposable tensor.

I have some ideas but none of them work, such as:

$$U = Im(\varphi) \simeq \frac{V_1 \times \cdots \times V_m}{Ker(\varphi)}$$

which is wrong because $\varphi$ is multilinear and not linear, so I have no idea how to solve this problem.

Any ideas? Hints? Thanks!

If you know where I can find this problem (maybe in some book) I would be very grateful.

• Your last equation doesn't make sense because $\varphi$ is not a homomorphism. – Matt Samuel Jan 30 '15 at 18:48
• $V_1 \times \cdots \times V_m$ and $U$ are vector spaces and $\varphi$ is linear (i.e. is a homomorphism of $\mathbb{K}$-modules (or $\mathbb{K}$-vector spaces), so I can apply the homomorphism theorem. Am I wrong? – Leafar Feb 1 '15 at 17:01
• The part that is wrong: $\varphi$ is not linear, it is multilinear. – Matt Samuel Feb 1 '15 at 17:02
• Ok, I will erase then – Leafar Feb 1 '15 at 17:04
• As written the problem has a gap: the "Show that..." part makes no use of $\varphi$, so $\varphi$ doesn't look relevant to the problem. You could give a stupid answer like $K = V_1 \otimes \cdots \otimes V_n$, for which the quotient space is zero. – KCd Feb 1 '15 at 17:53

By the universal property of the tensor product, $\varphi$ induces a linear map $\psi\colon V_1 \otimes \cdots \otimes V_m \to U$ which is also surjective (why?). And in fact we know more: for any $u \in U$, because $\varphi$ is surjective there are $v_i \in V_i$ such that $\varphi(v_1, \dots, v_m) = \psi(v_1 \otimes \cdots \otimes v_m) = u$. I'm hoping that from here you can fill in a gap. Thinking about the first isomorphism theorem is a good idea, though I don't think you need to apply it directly.

• Thanks! I will think about it. Actually I also tried to use the universal factorization property of $\otimes$. I am going to try again then... – Leafar Feb 1 '15 at 17:22
• Ok, we have $U \simeq \frac{V_1 \otimes \cdots \otimes V_m}{Ker(\psi)}$, so I let $K = Ker(\psi)$. We have $\frac{V_1 \otimes \cdots \otimes V_m}{Ker(\psi)} = \{v_1 \otimes \cdots \otimes v_m + Ker(\psi) : v_1 \otimes \cdots \otimes v_m \in V_1 \otimes \cdots \otimes V_m\}$ – Leafar Feb 2 '15 at 18:48
• We know that for any $u \in U$, $u$ is the image of a decomposable tensor, so each class of $\frac{V_1 \otimes \cdots \otimes V_m}{Ker(\psi)}$ is of the form $v_1 \otimes \cdots \otimes v_m + Ker(\psi)$. As $0 \in Ker(\psi)$, $v_1 \otimes \cdots \otimes v_m$ is in the class, so we can conclude that that each class of $\frac{V_1 \otimes \cdots \otimes V_m}{Ker(\psi)}$ contains a decomposable tensor. Right? – Leafar Feb 4 '15 at 2:24
• @Leafar Yes, that sounds right. – Hoot Feb 4 '15 at 18:26

By the universal factorization property of $\otimes$, there exists a unique linear map $\psi: V_1 \otimes \cdots \otimes V_m \longrightarrow U$ such that $\varphi = \psi \circ \otimes$.

As $\varphi$ is surjective, $\psi$ is surjective: given $u \in U$, $u = \varphi(v_1,\ldots,v_m)$ for some $(v_1,\ldots,v_m) \in V_1 \times \cdots \times V_m$, then $u = \psi \circ \otimes (v_1,\ldots,v_m) = \psi(v_1 \otimes \cdots \otimes v_m)$, then $u \in Im(\psi)$.

So given $u \in U$, there exists $(v_1,\ldots,v_m) \in V_1 \times \cdots \times V_m$ such that \begin{align*} u = \psi(v_1 \otimes \cdots \otimes v_m) = \varphi(v_1,\ldots,v_m). \end{align*}

Let $K=Ker(\psi)$.

Then, \begin{align*} \pi: \frac{V_1 \otimes \cdots \otimes V_m}{K} &\to U\\ z + Ker(\psi) &\mapsto \psi(z) \end{align*} is an isomorphism (easy to check).

Let us prove that each class of $\frac{V_1 \otimes \cdots \otimes V_m}{K}$ contains a decomposable tensor.

If $z + K = K$ it is trivial because $0 \in K$ and $0$ is decomposable.

Suppose that there exists a class $z_0 + K \neq K$ such that $z_0 + K$ does not contain a decomposable tensor.

Then, it does not exist $z \in z_0 + K$ decomposable such that $\pi(z + K) = \psi(z) = u$, but this is false, since each $u \in U$ is of the form $\psi(v_1 \otimes \cdots \otimes v_m)$ for some $(v_1,\ldots,v_m) \in V_1 \times \cdots \times V_m$ as we saw above.

So we conclude that each class of $\frac{V_1 \otimes \cdots \otimes V_m}{K}$ contains a decomposable tensor.