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Let $B_1, B_2$ be unit balls in finite-dimensional normed spaces $X_1, X_2$ respectively. Let $e(B_1), e(B_2)$ be corresponding extreme points sets.

Consider the unit ball $B$ in tensor product $X_1\otimes X_2$ with the largest (projective) cross-norm on it.

Can we say that extreme points of $B$ in tensor product are exactly tensor products of extreme points for $B_1, B_2$, i.e. $e(B)=\{u\otimes w: u\in e(B_1), w\in e(B_2)\}$?

This seems plausible, but things are not looking very straightforward. In particular, opposit pairs of extreme points produce the same point in tensor product, i.e. $(-u)\otimes (-w) = u\otimes w$.

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3 Answers 3

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See

[11] Ruess, W.M. and Stegall, C.P., Extreme points in duals of operator spaces, Math. Ann., 261 (1982), 535–546.

They prove what you want in a more general context: If $X$, $Y$ are Banach spaces s.t. either $X^*$ or $Y^*$ has the approximation property and either $X^*$ or $Y^*$ has the Radon-Nikodyn property, then the extreme points of the unit ball of the projective tensor product of $X^*$ and $Y^*$ are the tensor products of extreme points of the respective unit balls.

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  • $\begingroup$ I think Ruess, Stegall in their paper talk about the extreme point of the injective tensor product. How does the projective case follow from there? $\endgroup$
    – Topology
    Commented Feb 7, 2020 at 17:16
  • $\begingroup$ @Topology no, their paper deals with the dual of the injective tensor product. Bill Johnson's assumptions on $X$ and $Y$ ensure that $X^* \mathbin{\tilde\otimes_\pi} Y^* = (X \mathbin{\tilde\otimes_\varepsilon} Y)^*$ isometrically (see e.g. Theorem 16.6 in Defant and Floret, Tensor Norms and Operator Ideals, North-Holland, 1993). [...] $\endgroup$
    – J. van Dobben de Bruyn
    Commented Sep 29, 2020 at 19:49
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    $\begingroup$ [...] The injective unit ball does not preserve extreme points. Indeed, if $\mathcal H$ is a Hilbert space with $2 \leq \dim(\mathcal H) < \infty$, then $\mathcal H \mathbin{\otimes_\varepsilon} \mathcal H \cong B(\mathcal H)$ isometrically. But the extreme points of the closed unit ball of $B(\mathcal H)$ are the unitary matrices, whereas the pure tensors $x \mathbin{\otimes} y$ correspond with matrices of rank $\leq 1$. So in fact $\text{ext}(B_{\mathcal H \mathbin{\otimes_\varepsilon} \mathcal H})$ is disjoint from $\text{ext}(B_{\mathcal H}) \mathbin{\otimes} \text{ext}(B_{\mathcal H})$. $\endgroup$
    – J. van Dobben de Bruyn
    Commented Sep 29, 2020 at 19:49
  • $\begingroup$ Incidentally, Ruess and Stegall point out that this special case of their result had already been settled by I.I. Tseitlin, The extreme points of the unit ball of certain spaces of operators, Matematicheskie Zametki, vol. 20 (1976), issue 4, pp. 521–527. $\endgroup$
    – J. van Dobben de Bruyn
    Commented Sep 29, 2020 at 19:50
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Let me add another answer from the perspective of linear algebra/convex geometry. I think the real issue here is that the projective closed unit ball in the algebraic tensor product $E \mathbin{\otimes} F$ is, in general, not the convex hull of $B_E \mathbin{\otimes} B_F$, but rather its closed convex hull. If we omit the closure, even higher-dimensional faces are preserved:

Theorem. ([Dob20, Theorem 3.25]¹) Let $E$ and $F$ be real vector spaces, let $C \subseteq E$, $D \subseteq F$ be symmetric convex sets, and let $M \subset C$, $N \subset D$ be proper faces. Then $\text{conv}(M \mathbin{\otimes} N)$ is a face of $\text{conv}(C \mathbin{\otimes} D)$.

¹: Full disclosure: I am advertising my own paper here. The preceding theorem might have been known before, but I have not been able to find an earlier reference for this result.

In finite-dimensional spaces, $\text{conv}(B_E \mathbin{\otimes} B_F)$ is already closed (by a compactness argument), so here it follows that the projective closed unit ball preserves proper faces. (For extreme points, this also follows from Bill Johnson's answer.)

In general, the preceding theorem does not say anything about the projective unit ball, since an extreme point of $\text{conv}(B_E \mathbin{\otimes} B_F)$ is not necessarily an extreme point of its closure $\overline{\text{conv}}(B_E \mathbin{\otimes} B_F)$.

The other answers suggest that stronger assumptions are needed to ensure preservation of extreme points. I would still be interested in a counterexample to your question, but I guess Bill Johnson's answer suffices for most practical purposes.

References.

[Dob20]: J. van Dobben de Bruyn, Tensor products of convex cones, part I: mapping properties, faces, and semisimplicity, preprint (2020), arXiv:2009.11836v1.

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There is a hierarchy of notions related to the notion of extreme point, among them strong extreme point, point of continuity, denting point, as sketched in

BOR-LUH LIN, PEI-KEE LIN AND S. L. TROYANSKI. CHARACTERIZATIONS OF DENTING POINTS. PROC. OF AMS. Volume 102, Number 3, March 1988, p.526-528 (http://www.ams.org/journals/proc/1988-102-03/S0002-9939-1988-0928972-1/S0002-9939-1988-0928972-1.pdf).

The corresponding statement is true for denting points as shown in

DIRK WERNER. DENTING POINTS IN TENSOR PRODUCTS OF BANACH SPACES. PROC. OF AMS. Volume 101, Number 1, September 1987, p. 122-126 (http://www.ams.org/journals/proc/1987-101-01/S0002-9939-1987-0897081-1/S0002-9939-1987-0897081-1.pdf).

Now, it's enough to note that notions of extreme point and denting point are equivalent in my context.

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