projective tensor product

In functional analysis, an area of mathematics, the projective tensor product of two locally convex topological vector spaces is a natural topological vector space structure on their tensor product. Namely, given locally convex topological vector spaces $X$ and $Y$, the projective topology, or π-topology, on $X \otimes Y$ is the strongest topology which makes $X \otimes Y$ a locally convex topological vector space such that the canonical map $(x,y) \mapsto x \otimes y$ (from $X\times Y$ to $X \otimes Y$) is continuous. When equipped with this topology, $X \otimes Y$ is denoted $X \otimes_\pi Y$ and called the projective tensor product of $X$ and $Y$. It is a particular instance of a topological tensor product.

Definitions

Let $X$ and $Y$ be locally convex topological vector spaces. Their projective tensor product $X \otimes_\pi Y$ is the unique locally convex topological vector space with underlying vector space $X \otimes Y$ having the following universal property:
:For any locally convex topological vector space $Z$, if $\Phi_Z$ is the canonical map from the vector space of bilinear maps $X\times Y \to Z$ to the vector space of linear maps $X \otimes Y \to Z$, then the image of the restriction of $\Phi_Z$ to the ''continuous'' bilinear maps is the space of ''continuous'' linear maps $X \otimes_\pi Y \to Z$.
When the topologies of $X$ and $Y$ are induced by seminorms, the topology of $X \otimes_\pi Y$ is induced by seminorms constructed from those on $X$ and $Y$ as follows. If $p$ is a seminorm on $X$, and $q$ is a seminorm on $Y$, define their ''tensor product'' $p \otimes q$ to be the seminorm on $X \otimes Y$ given by
$(p \otimes q)(b) = \inf_ r$
for all $b$ in $X \otimes Y$, where $W$ is the balanced convex hull of the set $\left\$. The projective topology on $X \otimes Y$ is generated by the collection of such tensor products of the seminorms on $X$ and $Y$.
When $X$ and $Y$ are normed spaces, this definition applied to the norms on $X$ and $Y$ gives a norm, called the ''projective norm'', on $X \otimes Y$ which generates the projective topology.

Properties

Throughout, all spaces are assumed to be locally convex. The symbol $X \widehat_\pi Y$ denotes the completion of the projective tensor product of $X$ and $Y$.
* If $X$ and $Y$ are both Hausdorff then so is $X \otimes_\pi Y$; if $X$ and $Y$ are Fréchet spaces then $X \otimes_\pi Y$ is barelled.
* For any two continuous linear operators $u_1 : X_1 \to Y_1$ and $u_2 : X_2 \to Y_2$, their tensor product (as linear maps) $u_1 \otimes u_2 : X_1 \otimes_\pi X_2 \to Y_1 \otimes_\pi Y_2$ is continuous.
* In general, the projective tensor product does not respect subspaces (e.g. if $Z$ is a vector subspace of $X$ then the TVS $Z \otimes_\pi Y$ has in general a coarser topology than the subspace topology inherited from $X \otimes_\pi Y$).
* If $E$ and $F$ are complemented subspaces of $X$ and $Y,$ respectively, then $E \otimes F$ is a complemented vector subspace of $X \otimes_\pi Y$ and the projective norm on $E \otimes_\pi F$ is equivalent to the projective norm on $X \otimes_\pi Y$ restricted to the subspace $E \otimes F$. Furthermore, if $X$ and $F$ are complemented by projections of norm 1, then $E \otimes F$ is complemented by a projection of norm 1.
* Let $E$ and $F$ be vector subspaces of the Banach spaces $X$ and $Y$, respectively. Then $E \widehat F$ is a TVS-subspace of $X \widehat_\pi Y$ if and only if every bounded bilinear form on $E \times F$ extends to a continuous bilinear form on $X \times Y$ with the same norm.

Completion

In general, the space $X \otimes_\pi Y$ is not complete, even if both $X$ and $Y$ are complete (in fact, if $X$ and $Y$ are both infinite-dimensional Banach spaces then $X \otimes_\pi Y$ is necessarily complete). However, $X \otimes_\pi Y$ can always be linearly embedded as a dense vector subspace of some complete locally convex TVS, which is generally denoted by $X \widehat_\pi Y$.

The continuous dual space of $X \widehat_\pi Y$ is the same as that of $X \otimes_\pi Y$, namely, the space of continuous bilinear forms $B(X, Y)$.

Grothendieck's representation of elements in the completion

In a Hausdorff locally convex space $X,$ a sequence $\left(x_i\right)_^$ in $X$ is absolutely convergent if $\sum_^ p \left(x_i\right) < \infty$ for every continuous seminorm $p$ on $X.$ We write $x = \sum_^ x_i$ if the sequence of partial sums $\left(\sum_^n x_i\right)_^$ converges to $x$ in $X.$

The following fundamental result in the theory of topological tensor products is due to Alexander Grothendieck.

The next theorem shows that it is possible to make the representation of $z$ independent of the sequences $\left(x_i\right)_^$ and $\left(y_i\right)_^.$

Topology of bi-bounded convergence

Let $\mathfrak_X$ and $\mathfrak_Y$ denote the families of all bounded subsets of $X$ and $Y,$ respectively. Since the continuous dual space of $X \widehat_\pi Y$ is the space of continuous bilinear forms $B(X, Y),$ we can place on $B(X, Y)$ the topology of uniform convergence on sets in $\mathfrak_X \times \mathfrak_Y,$ which is also called the topology of bi-bounded convergence. This topology is coarser than the strong topology on $B(X, Y)$, and in , Alexander Grothendieck was interested in when these two topologies were identical. This is equivalent to the problem: Given a bounded subset $B \subseteq X \widehat Y,$ do there exist bounded subsets $B_1 \subseteq X$ and $B_2 \subseteq Y$ such that $B$ is a subset of the closed convex hull of $B_1 \otimes B_2 := \$?

Grothendieck proved that these topologies are equal when $X$ and $Y$ are both Banach spaces or both are DF-spaces (a class of spaces introduced by Grothendieck). They are also equal when both spaces are Fréchet with one of them being nuclear.

Strong dual and bidual

Let $X$ be a locally convex topological vector space and let $X^$ be its continuous dual space. Alexander Grothendieck characterized the strong dual and bidual for certain situations:

Examples

*For $(X, \mathcal, \mu)$ a measure space, let $L^1$ be the real Lebesgue space $L^1(\mu)$; let $E$ be a real Banach space. Let $L^1_E$ be the completion of the space of simple functions $X\to E$, modulo the subspace of functions $X\to E$ whose pointwise norms, considered as functions $X\to\Reals$, have integral $0$ with respect to $\mu$. Then $L^1_E$ is isometrically isomorphic to $L^1 \widehat_\pi E$.

See also

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Citations

References

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Further reading

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External links

Nuclear space at ncatlab

{{Functional analysis

Functional analysis
Topological tensor products