Added tensor gymnastics.

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Bokuan Li
2026-07-20 12:58:27 -04:00
parent 4b1a17c259
commit 22f9bc219b

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@@ -53,6 +53,32 @@
In constructing the \hyperref[projective tensor product]{definition:projective-tensor-product}, it may be more natural to obtain its topology as a projective topology using its universal property. However, doing so requires taking a least upper bound across \textit{all continuous linear maps defined on} $E \times F$, a collection too big to be a set. As such, constructing it as a projective topology is logically dubious, or at the very least beyond my abilities.
\end{remark}
\begin{proposition}
\label{proposition:projective-tensor-product-dual}
Let $E, F$ be locally convex space over $K \in \RC$, then
\[
(E \wh \otimes_\pi F)^* = (E \otimes_\pi F)^* \iso L^2(E, F; K) \iso L(E; F^*) \iso L(F; E^*)
\]
where:
\begin{enumerate}
\item The dual pairing between $E \otimes_\pi F$ and $L^2(E, F; K)$ is given by
\[
\angles{\sum_{k = 1}^n x_k \otimes y_k, \lambda}_{E \otimes_\pi F} = \sum_{k = 1}^n \lambda(x_k, y_k)
\]
\item The dual pairing between $E \otimes_\pi F$ and $L(E; F^*)$ is given by
\[
\angles{\sum_{k = 1}^n x_k \otimes y_k, T}_{E \otimes_\pi F} = \sum_{k = 1}^n \dpn{y_k, Tx_k}{F}
\]
\item The dual pairing between $E \otimes_\pi F$ and $L(F; E^*)$ is given by
\[
\angles{\sum_{k = 1}^n x_k \otimes y_k, T}_{E \otimes_\pi F} = \sum_{k = 1}^n \dpn{x_k, Ty_k}{E}
\]
\end{enumerate}
\end{proposition}
\begin{definition}[Cross Seminorm]