5
$\begingroup$

As presented in the answer of this post, the projective tensor norm on the algebraic tensor product of two Banach spaces $X$ and $Y$ is given by \[ \Vert \omega\Vert_{\pi} = \inf\left\{\sum \lVert x_{i}\rVert_X \,\lVert y_{i}\rVert_ Y\,:\,\omega = \sum_{i=1}^{n} x_{i} \otimes y_{i}\right\} \] with respect to which we complete the algebraic tensor product to obtain the tensor product Banach space $X \otimes Y$

For Hilbert spaces $H_1$ and $H_2$, as shown in this link, the inner product on the algebraic tensor product is given by linear extension of the formula \begin{equation} \langle \phi_1 \otimes \phi_2, \psi_1 \otimes \psi_2 \rangle = \langle \phi_1, \psi_1 \rangle_{H_1} \langle \phi_2, \psi_2 \rangle_{H_2} \end{equation} for $\phi_1, \psi_1 \in H_1$ and $\phi_2, \psi_2 \in H_2$. Next, the tensor product Hilbert space $H_1 \otimes H_2$ is defined to be completion under this inner product.

Now, my question is

Does the projective tensor norm on general Banach spaces, as defined above for $X$ and $Y$, imply the inner product $\langle \phi_1 \otimes \phi_2, \psi_1 \otimes \psi_2 \rangle$ when $X$ and $Y$ are both Hilbert spaces?

I think this must be the case, but it does not seem obvious at all. Could anyone please provide any clarification?

$\endgroup$
2
  • 2
    $\begingroup$ It defenitely not the case. The Hilbert tensor product gives a Hilbert space. Meanwhile the the projective tensor product is rarely reflexive not (to mention Hilbertable). $\endgroup$
    – Norbert
    Mar 1 at 1:11
  • $\begingroup$ @Norbert I see..Isn't it at least true that $\lVert x \otimes y \rVert_{X \otimes Y} =\lVert x \rVert_X \lVert y \rVert_Y$? $\endgroup$
    – Keith
    Mar 1 at 8:37

2 Answers 2

6
$\begingroup$

The Hilbert tensor product is in general not equal to the projective tensor product:

If $H$ is a Hilbert space and $H^*$ its dual space, then

  • $H \hat \otimes_\pi H^*$ (the projective tensor product) is (isometrically isomorphic to) the trace class (nuclear) operators with the trace norm
  • $H \hat \otimes_\epsilon H^*$ (the injective tensor product) is (isometrically isomorphic to) the compact operators with the operator norm
  • $H \hat \otimes_h H^*$ (the Hilbert tensor product, which is a Hilbert space again) is (isometrically isomorphic to) the Hilbert-Schmidt operators with the Hilbert-Schmidt norm

Since the spaces of Hilbert-Schmidt, compact and trace class operators are, in infinite dimensions, never the same it follows that the tensor norms can not be the same either.

To show that they are never the same in infinite dimensions:

Let $(e_n)_{n \in \mathbb{N}}$ be an orthonormal system in $H$. Let $(x_n)_{n \in \mathbb{N}}$ be any sequence of complex numbers that converges to 0, but whose absolute value is not square summable. For example $x_n =1/\sqrt{n}$. Define a linear operator $T: H\to H$ by $Ty = \sum_{n \in \mathbb{N}} x_n \langle e_n, y \rangle e_n$. Then $T$ is compact, but not Hilbert-Schmidt.

Let $(s_n)_{n \in \mathbb{N}}$ be any sequence of complex numbers whose absolute value is square summable, but not summable. For example $s_n := 1/n$. Define a linear operator $S : H \to H$ by $Sy = \sum_{n \in \mathbb{N}} s_n \langle e_n, y \rangle e_n$. Then $S$ is Hilbert-Schmidt, but not trace class.

$\endgroup$
10
  • 1
    $\begingroup$ @Keith yes, that is true for every "well behaved" norm on the tensor product (in particular this is true for all three that are discussed here). $\endgroup$
    – jd27
    Mar 1 at 8:55
  • 1
    $\begingroup$ @Keith the complexity arrives by the way the norm is defined for a general tensor of the form $\sum_{i=1}^n x_i \otimes y_i$. All 3 norms in discussion do this differently. $\endgroup$
    – jd27
    Mar 1 at 8:58
  • 1
    $\begingroup$ @Keith It is unclear to me what $\| \cdot \|$ is refering to in your comment! $\endgroup$
    – jd27
    Mar 1 at 9:07
  • 1
    $\begingroup$ @Keith Yes it is true, which is exactly what my response comment to your first comment asserts. It was unclear to me that you were talking about the projective tensor norm and not about an arbitrary norm in your initial comment. $\endgroup$
    – jd27
    Mar 1 at 9:13
  • 1
    $\begingroup$ @Keitch The resulting Banach space norm is also never equivalent to the Hilbert space norm if both of the factors are infinite dimensional (see the answer by Norbert) $\endgroup$
    – jd27
    Mar 1 at 13:23
4
$\begingroup$

Suppose $X$ and $Y$ are Hilbert spaces and $X\hat\otimes_\pi Y$ is isomorphic to $X\hat\otimes_h Y$. Since the latter space is reflexive, then so is $X\hat\otimes_\pi Y$. From theorem 4.21 in Introduction to tensor products of Banach spaces. R.A. Ryan we know that $X\hat\otimes_\pi Y$ is reflexive iff every operator operator from $X$ to $Y^*$ is compact. For Hilbert spaces $X$ and $Y$ this is the case only if $X$ or $Y$ is finite dimensional.

I think that one of the spaces has to be at most one dimensional if we require isometric isomorphism between $X\hat\otimes_\pi Y$ and $X\hat\otimes_h Y$.

$\endgroup$
3
  • 1
    $\begingroup$ If I recall correctly, in the case of Hilbert spaces the injective tensor product is isometrically isomorphic to the space of compact operators from $X^\ast$ to $Y$, so it can never be isometrically isomorphic to a Hilbert space unless either $X$ or $Y$ is one-dimensional (or $0$, of course). I don't know about projective tensor products though. $\endgroup$
    – David Gao
    Mar 1 at 3:28
  • 1
    $\begingroup$ Also, a small correction: every operator from $X$ to $Y^\ast$ is compact iff either $X$ or $Y$ is finite-dimensional, not specifically that $Y$ is finite-dimensional. $\endgroup$
    – David Gao
    Mar 1 at 3:30
  • $\begingroup$ @DavidGao, thank you. Fixed $\endgroup$
    – Norbert
    Mar 1 at 21:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .