2
$\begingroup$

Let $X_1, X_2, Y_1, Y_2$ be $\mathbb{C}$-vector spaces and $f_1: X_1 \to Y_1$ and $f_2: X_2 \to Y_2$ conjugate-linear maps, i.e. $f_1(\alpha x_1) = \overline{\alpha} f_1(x_1)$ and $f_1(x_1 + x_1') = f_1(x_1) + f_1(x_1')$ and similarly for $f_2$.

Question: Does there exist a unique conjugate linear map $$f_1 \otimes f_2: X_1 \otimes X_2 \to Y_1 \otimes Y_2$$ such that $(f_1 \otimes f_2)(x_1 \otimes x_2) = f_1(x_1) \otimes f_2(x_2)?$

Basically, we would want to apply the universal property of the tensor product to construct this map, but the maps are not linear. But I believe there should be a trick to reduce the problem to linear maps.

An alternative way of approaching this: let $\{e_i\}_{i \in I}$ be a basis for $X_1$ and define $$(f_1\otimes f_2) (\sum_{i \in I} e_i \otimes z_i) := \sum_{i \in I} f_1(e_i) \otimes f_2(z_i)$$ and this gives existence of the map, but it still looks like an unnatural way to proceed.

$\endgroup$
0
5
$\begingroup$

If $V$ is a $\Bbb C$-vector space, then we can define a conjugate vector space $c(V)$ as follows.

The underlying abelian group of $c(V)$ is the same as the abelian group $V$. The scalar product $\bullet$ in $c(V)$ is defined as $\alpha \bullet v = \overline\alpha \cdot v$, where $\cdot$ denotes the scalar product on $V$.

Thus by definition, the map $f_1$ is nothing but a $\Bbb C$-linear map from $X_1$ to $c(Y_1)$. Same for $f_2$.

Now it only remains to prove a

Lemma: $c(Y_1) \otimes c(Y_2)$ is canonically isomorphic to $c(Y_1 \otimes Y_2)$ as $\Bbb C$-vector spaces.

The proof of the lemma is a simple exercise, and the original question immediately follows from the lemma.

$\endgroup$
0

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.