$ L^2 (\mathbb{R}, \mathbb{C}) \otimes \mathbb{C} ^2 $ is the same as $L^2(\mathbb{R}, \mathbb{C}) \times L^2(\mathbb{R}, \mathbb{C})$

Question

I want to show that $ L^2 (\mathbb{R}, \mathbb{C}) \otimes \mathbb{C} ^2 $ can be thought of as the same space as $L^2(\mathbb{R}, \mathbb{C}) \times L^2(\mathbb{R}, \mathbb{C})$ Perhaps they are isometrically isomorphic.

What is causing me trouble is the fact that we have an infinite-dimensional space here. Perhaps I could take an orthonormal Schauder basis for $ L^2 $, but the fact that these spaces are the same feel like it should be obvious.

My question is, is it correct that $ L^2 (\mathbb{R}, \mathbb{C}) \otimes \mathbb{C} ^2 $ can be thought of as the same space as $L^2(\mathbb{R}, \mathbb{C}) \times L^2(\mathbb{R}, \mathbb{C})$? If so, how would one prove it?

Answer 1

You can see this in many ways, but one is through properties of the tensor product:

For modules $M,N_1,$ and $N_2$ over a commutative ring $R$ $$M\otimes (N_1\oplus N_2)=(M\otimes N_1)\oplus (M\otimes N_2)$$ and $$M\otimes R=M$$ see Lemma 8.12 here or any text covering tensor products.

In your example, $M$ is $L^2(\mathbb{R},\mathbb{C})$ and $N_1$, $N_2$ , and $R$ are all $\mathbb{C}$. From these two general properties you can deduce the desired isomorphism of $\mathbb{C}$-modules, that is vector spaces over $\mathbb{C}$.

This is a little abstract so if it helps you can think of it this way: each element of $L^2(\mathbb{R},\mathbb{C})\otimes \mathbb{C}^2$ is a sum of elements of the form $f\otimes(z_1,z_2)$, which using the properties of the tensor product may be rewritten as $z_1f\otimes(1,0)+z_2f\otimes(0,1)$. Thus, if $\{f_i\}$ is a basis of $L^2(\mathbb{C},\mathbb{R})$, $\{f_1\otimes(1,0)\}\cup \{f_i\otimes(0,1)\}$ is a basis of $L^2(\mathbb{R},\mathbb{C})\otimes \mathbb{C}^2$.

Answer 2

Define $V:L^2 (\mathbb{R}, \mathbb{C}) \otimes \mathbb{C} ^2\to L^2 (\mathbb{R}, \mathbb{C}) \oplus_2 L^2 (\mathbb{R}, \mathbb{C}) $ in the following way. Start with $$ V(f\otimes (a,b))=af\oplus bf. $$ Extend by linearity to the linear span of a basis. Now, if $f_1,\ldots,f_n$ are orthonormal, then \begin{align} \Big\|V\,\Big[\sum_{k=1}^nc_kf_k\otimes (1,0)+\sum_{j=1}^nd_jf_j\otimes(0,1)\Big]\Big\|^2 &=\Big\|\,\sum_{k,j=1}^nc_k f_k\oplus d_jf_j\Big\|^2\\[0.3cm] &=\sum_{k,j=1}^n|c_k|^2+|d_j|^2\\[0.3cm] &=\Big\|\sum_{k=1}^nc_kf_k\otimes (1,0)+\sum_{j=1}^nd_jf_j\otimes(0,1)\Big\|^2. \end{align} So $V$ is isometric, which allows us to extend it to all of $L^2(\mathbb R,\mathbb C)\otimes\mathbb C^2$. Its range is dense and closed, so $V$ is surjective. That is, $V$ is an isometric isomorphism.

Answer 3

So far the answers given have directed attention towards "complicated things" like what is the tensor product for infinite-dimensional vector spaces. Let me offer another answer based on the totally opposite viewpoint of focusing on "simple things" like asking what is $\mathbb{C}^2$?

Well, $\mathbb{C}^2=L^2(\{1,2\},\mathbb{C})$, i.e., it is the space of complex-valued functions on the finite set $X=\{1,2\}$. One could also put on it the $\sigma$-algebra $\{\varnothing,\{1\},\{2\},\{1,2\}\}$ and the counting measure and then there would be no cheating when viewing $\mathbb{C}^2$ as an honest $L^2$ space.

Now the raison d'être of tensor products is to correspond to functions on the Cartesian product when the individual vector spaces are presented to us as spaces of functions. So $$ L^2(\mathbb{R},\mathbb{C})\otimes \mathbb{C}^2=L^2(\mathbb{R},\mathbb{C})\otimes L^2(\{1,2\},\mathbb{C})=L^2(\mathbb{R}\times\{1,2\},\mathbb{C}) $$ and elements of the last space trivially correspond to pairs of functions $f_1(x)$, $f_2(x)$ in $L^2$.