4
$\begingroup$

If $U$ and $V$ are finite-dimensional vector spaces then $U^*\otimes V^* \approx (U \otimes V)^*$ via the isomorphism $\tau: U^*\otimes V^* \to(U \otimes V)^*$ given by $\tau(f \otimes g)(u \otimes v) = f(u)g(v)$.

And more generally the linear transformation $\theta: \mathcal{L}(U,U') \otimes \mathcal{L}(V,V') \to \mathcal{L}(U \otimes V, U' \otimes V')$ defined by $\theta(\tau \otimes \sigma) = \tau \odot \sigma$ where $(\tau \otimes \sigma)(u \otimes v) = \tau(u) \otimes \sigma(v)$ is an injective linear transformation and is an isomorphism if all vector spaces are finite-dimensional.

In the proof of these what exactly goes wrong for infinite dimensional vector spaces so that the linear transformation is not surjective.

An example showing non-surjective would also be appreciated.

$\endgroup$
1
$\begingroup$

Let $V$ be an infinite dimensional vector space over $k$. Let $W$ be the image of

$$\mathcal L(V, k)\otimes \mathcal L(k,V) \to \mathcal L(V \otimes k, k\otimes V).$$

Let us identify these objects and this map with

$$V^\vee \otimes V \to \mathcal L(V, V).$$

Suppose that $f : V \to V$ is in $W$. Show that $f(V)$ is finite-dimensional. In particular, the identity map of $V$ is not in $W$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.