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I have two questions which are intuitively true.

Let $V$ be a Hilbert space. As usual we can turn $V\otimes V$ or $V\otimes V\otimes V$ into Hilbert spaces by intorducing the natural inner product and by performing completion.

Question 1. We have a sequence of simple tensors $f_{i}\otimes g_{i}$ that converges in the Hilbert space $V\otimes V$ to some tensor. Is it true that the limit is a simple tensor, that is can be represented by $f\otimes g$? The matter is that $f_{i}$ and $g_{i}$ need not have limits, as the simple example $i f\otimes\frac{1}{i} g$, for some fixed $f,g\in V$ shows.

Question 2. If a sequence of simple tensors of the form $f_{i}\otimes f_{i}\otimes g_{i}$ has a limit which is a simple tensor, can it be represented by $f\otimes f\otimes g$, for some $f,g\in V$?

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    $\begingroup$ For the i * 1/i problem, probably you should take a hint from algebraic geometry and projectivize your spaces. Then your question is asking if the "Segre variety" is still closed for projective Hilbert spaces instead of usual projective spaces. $\endgroup$ – John Wiltshire-Gordon Sep 21 '12 at 14:43
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Under the natural identification of the completion of $V\otimes W$ with Hilbert-Schmidt operators $V\rightarrow W^*$, monomial tensors give rank-one operators. A Hilbert-Schmidt-norm limit of rank-one operators is certainly rank-one. (This viewpoint gets away from the pitfalls of specific representations of the tensors.) A similar argument works for $V\otimes V\otimes V$.

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It is perhaps worth putting on record that this is a special case of an elementary result. A function $f$ on the product $X \times Y$ of two abstract sets is a simple tensor if and only if it is the product $f(x,y) = g(x)h(y)$ of two functions of one variable. Since this can be expressed in the pointwise condition $f(x_1,y_1)f(x_0,y_0)=f(x_1,y_0)f(x_0,y_1)$ for all suitable pairs, the class of simple tensors is closed in most useful topologies. The versions of this result which are useful in linear algebra or functional analysis can be obtained from this one by using the standard method of regarding vectors as functions on the dual space. If $f$ is continuous, then so are $g$ and $h$. If it is bilinear, then $g$ and $h$ are linear.

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The set of simple tensors in a tensor product of vector spaces $V^{\otimes k}$ is the (cone over) the Segre variety, $X$, which is a closed set if you include the zero vector. $X$ is also homogeneous -- it is the orbit under $GL(V)^k$ of a single vector $e\otimes \dots \otimes e$. In particular every point of $X$ is either a simple tensor or 0.

For Question 2, the set of tensors of the form $f\otimes f \otimes g$ is a Segre-Veronese variety (when you include 0), and it is also closed and homogeneous, so it contains all its limit points and all points are of the same form (or 0).

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