3
$\begingroup$

Let $H_1$ be a $\mathbb{C}$-Hilbert space and $H_2$ be a $\mathbb{R}$-Hilbert space, both assumed separable.

For any $\phi\in H_1,\psi\in H_2$ we may define the (bilinear? by the wikipedia definition this can't be bilinear since $H_1$ and $H_2$ has different scalar fields) mapping $\phi\otimes \psi:H_1\times H_2 \to \mathbb{C}$ by $$ \phi\otimes \psi(x,y) = \langle x,\phi \rangle_1 \langle y,\psi \rangle_2 $$ We note that since only $H_1$ has a complex inner product, we can in general for $\lambda\in \mathbb{C}$ only say that $$ \lambda (\phi \otimes \psi) = (\lambda \psi)\otimes \psi \quad \quad \text{and not}\quad \quad \lambda (\phi \otimes \psi) = \psi\otimes (\lambda\psi). $$ Anyways I have proven (I can add the proof, but i think it is correct) that we can create an inner product of the space $\mathcal{E}$ of all finite linear combinations of the bilinear mappings considered above, by letting $$ \langle \phi_1 \otimes \psi_1 , \phi_2\otimes \psi_2 \rangle = \langle\phi_1,\phi_2 \rangle_1 \langle\psi_1,\psi_2 \rangle_2 $$ and extending it to finite linear combinations in the following way $$ \Big\langle \sum_{i=1}^n a_i ( \phi_i\otimes \psi_i), \sum_{i=1}^m b_i (\beta_i \otimes \gamma_i) \Big\rangle = \sum_{i=1}^n \sum_{j=1}^m a_i \bar{b}_j \langle \phi_i\otimes \psi_i, \beta_j \otimes \gamma_j \rangle $$ We now define $H_1\otimes H_2$ as the completion of the space of finite linear combinations with respect to the metric induced by the above inner product. Furhtermore it is well-known that the above inner product can be extended to $H_1\otimes H_2$, such that it satisfies $$ \langle \iota(\phi\otimes \psi), \iota(\gamma\otimes \beta) \rangle_{H_1\otimes H_2} = \langle\phi_1,\gamma\rangle_1 \langle \psi,\beta\rangle_2 $$ for any $\phi,\gamma\in H_1$ and $\psi,\beta\in H_2$, where $\iota$ is the linear isometric embedding into the completion.

Question: Is this a valid construction of the tensor product of two Hilbert spaces with different fields? I could not find any mistake, but the ideas are taken from some notes that considers two real Hilbert spaces. Also every source of the tensor product of Hilbert spaces that I have encountered considers either two real or two complex Hilbert spaces, which is why I'm worried I have made a mistake.

Edited to reflect only the above question remains.

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ You have some typos right at the beginning, where you are writing some things like $\psi\otimes\psi$. On the other hand your map $f$ is bilinear and $\|f(\phi,\psi)\|^2=\|\phi \otimes \psi \|^2 = \|\phi\|^2\|\psi\|^2$. So you have $\|f\|_\infty=1$ and $f$ is continuous. $\endgroup$ – s.harp Dec 7 '16 at 19:49
  • $\begingroup$ @s.harp But is it correct that I am allowed to take two Hilbert spaces with different scalar fields, and have a well-defined tensor product? In most resources on the internet one considers either two real or two complex Hilbert spaces, so I am not sure if I am allowed to do the above. $\endgroup$ – Martin Dec 8 '16 at 0:24
  • $\begingroup$ Consider first $H_1\otimes H_2$ where you understand $H_1, H_2$ as real vector spaces, then you notice that this has the structure of a complex vector space, namely on the pure tensors for complex $\lambda$: $$\lambda \cdot(v\otimes w) = v\otimes (\lambda w)$$ and scalar multiplication with complex numbers is well defined. I'm not entirely sure what this kind of a construction is a called, but it is related to extension/restriction of scalars. $\endgroup$ – s.harp Dec 8 '16 at 10:33
  • 1
    $\begingroup$ (Actually it is scalar extension: The construction is $H_1\otimes_{\mathbb C}(\mathbb C \otimes_{\mathbb R} H_2)$, here $\mathbb C\otimes_{\mathbb R} H_2$ is the scalar extension or complexification of the vector space $H_2$) $\endgroup$ – s.harp Dec 8 '16 at 13:32
  • $\begingroup$ @s.harp Thank you very much for the idea with the complexification of the non-complex space. Now it follows the standard approach seen in much litterature, but I'm still not quite sure why the above approch should fail. Anyways maybe write you your above comment as an answer and I will accept it. $\endgroup$ – Martin Dec 13 '16 at 2:25
1
$\begingroup$

Notice that $\mathbb C$ is a $\mathbb R$ vector-space, this means that any $\mathbb C$ vector-space can be viewed as an $\mathbb R$ vector-space, this is called restriction of scalars and there is a related construction called extension of scalars.

If $V$ is a $\mathbb R$ vector-space, then you can consider the tensor-product of real vector-spaces: $$\mathbb C\otimes_{\mathbb R}V$$ Now you also find that this has the structure of a complex vector-space, let $\lambda,z \in\mathbb C v\in V$ then $$\lambda\cdot(z\otimes_{\mathbb R}v)=(\lambda z)\otimes_{\mathbb R}v$$ extends to a well-defined scalar multiplication on all of $\mathbb C\otimes_{\mathbb R}V$. You are extending $V$ by the scalars you are missing in $\mathbb C$. This allows you to define the tensor-product of a $k_1$ vector-space $V$ with a $k_2$ vector-space $W$, where $k_1,k_2$ are fields and $k_2$ has a $k_1$ vector-space structure: $$V \otimes_{k_2} W := V\otimes_{k_2}(k_2\otimes_{k_1}W)$$ alternatively you could also look at $V$ as a $k_1$ vector-space and consider $V{\otimes}_{k_1}W$.

Your considerations about the tensor-products of Hilbert spaces work in this context.

|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.