The positive solution to the homogeneous space problem yields that l₂ is the only infinite-dimensional Banach space, up to isomorphism, which is isomorphic to all its infinite-dimensional subspaces. For a Banach space X which is not isomorphic to l₂, we investigate the problem of finding the number of non-isomorphic infinite-dimensional subspaces of X. As a consequence of our construction we also obtain a structural result about Banach spaces containing an unconditional basic sequence.

We say a multifunction T : X ® 2^X* is monotone provided that for all x,y Î X, and for all x^* Î T(x), y^* Î T(y),

In this talk, largely based on [1], I shall show how-thanks largely to a long-neglected observation of Fitzpatrick-the originally quite complex theory of monotone operators can be almost entirely reduced to convex analysis. I shall also highlight various long-standing open questions to which these new techniques have offered new access, [2], [3].

References

[1]

J. M. Borwein, Maximal Monotonicity via Convex Analysis. J. Convex Analysis (special issue in memory of Simon Fitzpatrick), 13(2006), June. [D-drive Preprint 281].

[2]

     , Maximality of Sums of Two Maximal Monotone Operators in General Banach Space. Proc. Amer. Math. Soc., accepted September 2006. [D-drive Preprint 322].

[3]

A. E. Eberhard and J. M. Borwein, Maximality of Monotone Operators in Banach Space. Preprint, September 2006. [D-drive Preprint 325].

The Morse-Sard theorem states that if f : Rⁿ ®R^m is a C^r smooth function, with r > max{n-m,0}, and C_f stands for the set of critical points of f (that is, the points of X at which the differential of f is not surjective), then the set of critical values, f(C_f), is of (Lebesgue) measure zero in R^m. This result is no longer true for functions defined on infinite dimensional Banach spaces. However, under this setting it is possible to develop a strong approximate version of such classical principle.

I was totally surprised that all of my earliest research in the 1970s has been used intensively in the recent development of quantum computers. By all means, I need to seek the new meanings of the old values of completely positive linear maps in the setup of non-commutative matrices.

The relative position of two 1-dimensional subspaces of Hilbert space is expressed fully by a single number in [0  1]. But suppose we study subspaces P spanned by orthonormal p₁, p₂, ..., p_k and Q spanned by orthonormal q₁, q₂, ..., q_k for higher k. To express the relative position of P and Q fully requires an unordered k-tuple of numbers; this theory has been understood since the 1880s and generalized to infinite-dimensional subspaces. In contrast, it is known that the relative position of three k-dimensional subspaces can not be expressed fully by a manageable invariant.

Nevertheless the manifold of k-subspaces can be studied to advantage. Here is a typical example of a question which can be asked of three subspaces and may have illuminating answers: Given subspaces P, Q, R, is P closer to Q than R is? This talk gives a modern way of dealing with such questions.

Let L(a) be an operator function of the class C¹ ([a,b], S(H) ) such that L(a) << 0, L(b) >> 0 and for all x Î H \{0} the function ( L(a)x, x ) has exactly one zero p(x) in (a,b). Define the following nonlinear operator

In 2002 Gluskin and I showed that a polytope with few vertices is far from being symmetric in a sense of Banach-Mazur distance. More precisely, it was shown that Banach-Mazur distance between such a polytope and any symmetric convex body is large. In this talk we introduce a new, averaging-type parameter to measure the asymmetry. It turns out that, surprisingly, this new parameter is still very large, in fact it satisfies the same lower bound as Banach-Mazur distance. We apply our results to provide a lower estimate on the vertex index of a symmetric convex body, which was recently introduced in by Bezdek and myself. Furthermore, we give the affirmative answer to a conjecture by K. Bezdek on the quantitative illumination problem.

This is joint work with E. D. Gluskin.

We investigate the properties and structure of topologically transitive multiplicative semigroups of real or complex matrices, and are particularly interested in the question: "What extra conditions must be imposed on such semigroups to guarantee transitivity?"

A set S of matrices is topologically transitive if any non-zero vector can be mapped arbitrarily close to any other vector by a matrix in S, and is transitive if any non-zero vector can be mapped extactly to any other vector by a matrix in S.

This talk is based on joint work with Leo Livshits and Heydar Radjavi.

I will discuss some results on irreducible semigroups of matrices that Heydar Radjavi and I have recently obtained, and consider which of these results might possibly generalize to semigroups of operators on Hilbert space.

We have established the following: a transitive, closed, homogeneous semigroup of linear transformations on a finite-dimensional space either has zero divisors or is simultaneously similar to a group consisting of scalar multiples of unitary transformations. The proof begins by establishing the result that for each closed homogeneous semi-group with no zero divisors there is a k such that the spectral radius of AB is at most kr(A) r(B) for all A and B in the semigroup. (It is also shown that the spectral radius is not k-submultiplicative on any transitive semigroup of compact operators.)

We have also proven that an irreducible semigroup of complex matrices is, respectively, finite, countable or bounded if there is a non-zero linear functional whose range on the semigroup has the corresponding property. Moreover, an irreducible semigroup is finite, countable or bounded if it has a nonzero ideal with the corresponding property.

I'll discuss as much of the above as time permits.

We investigate smooth points (points where the norm is differentiable) in C^*-algebras and von Neuman algebras. The goal is to obtain characterizations that connect to the operator algebra structure.

This talk is based on joint work with Wend Werner.