Let $\mathcal{H}$ be a Hilbert space (infinite dimensional, in general) then consider its tensor square: $\mathcal{H} \otimes \mathcal{H}$. Is the space $$\{f\otimes f \ | \ f\in \mathcal{H}\}$$ Dense in $\mathcal{H} \otimes \mathcal{H}$? If this is true then I suppose it is also true for arbitrary tensor powers of $\mathcal{H}$ also.
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Do you mean the set ${f\otimes f | f \in \mathcal H}$ or the linear span of this set? – lisyarus May 06 '19 at 08:56
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Did you try the case $\mathrm H = \mathbb R^2$ with the representation $\mathbb R^2\otimes \mathbb R^2 \cong \mathbb R^{2\times 2}$? – Jochen May 06 '19 at 11:23
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See the answer there for a more general question of whether the set of $f \otimes g$ with $f,g \in \mathcal H$, is dense (it's not). Note that I mean this set itself, not its linear span. https://math.stackexchange.com/questions/2397124/dense-subset-of-l2-of-product-space-with-separated-variables/2397350#2397350 – shalop May 06 '19 at 15:04
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I'll talk about the subspace $\mathcal L \subset \mathcal H \otimes \mathcal H$ spanned by $\{f\otimes f\}$.
$\mathcal L$ is not dense in $\mathcal H \otimes \mathcal H$ unless $\dim \mathcal H \leq 1$.
Consider the operator on $\mathcal H \otimes \mathcal H$ defined as
$$A(x \otimes y) = x\otimes y - y \otimes x$$
and extended by linearity. $A$ is easily seen to be continuous; furthemore, $\forall v \in \mathcal L \quad A(v) = 0$.
However, if $x,y \in \mathcal H$ are linearly independent, $A(x \otimes y) \neq 0$. Thus, $\ker A$ is a proper closed subspace of $\mathcal H \otimes \mathcal H$ containing $\mathcal L$, so the latter cannot be dense.
lisyarus
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I don't think that this is true. In fact, ${e_i \otimes e_i}$ does not need to be a basis of $\mathcal L$. Consider $\mathcal H = \mathbf R^2$. Then, $a\otimes a$, $b \otimes b$ and $c \otimes c$ with $a = (1,0)$, $b = (0,1)$ and $c = (1,1)$ are linearly independent. In particular, $\mathcal L$ is at least three-dimensional. – gerw May 06 '19 at 10:43
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