Gelfand representation

The Gelfand transform ( after Israel Gelfand ) is the main tool in the theory of commutative Banach algebras. It forms a commutative Banach algebra A - into an algebra of continuous functions. Each of A from A is assigned to a continuous function with a suitable locally compact Hausdorff space. The assignment is a continuous algebra homomorphism.

Motivation, Gelfand space

Consider a commutative Banach algebra A - only normed space with dual space and Bidualraum so, the elements of A as follows mapped to continuous functions: you arrange each a the function. It is the well-known isometric embedding of A in the Bidualraum because each is an item. Each is also continuous. Here, the norm topology is proving harder than necessary. For this reason, we consider the weak -* topology, this is just defined as the coarsest topology that makes all pictures continuously.

Let us return to the algebra A, then we must conclude that the association is not a homomorphism; it is not multiplicative, i.e. it does not apply. This would namely order to apply to all, but a linear functional is not multiplicative in general. This observation gives a clue as to how you can construct a homomorphism of the desired type. One uses only the very place multiplicative functionals in, and this is precisely the Gelfand transform.

Therefore we set. This amount is called the spectrum ( Gelfand ) spectrum A or the Gelfand area A. Note that the Nullhomomorphismus was taken out. There are Banach algebras with empty spectrum, for example, is a Banach algebra with zero multiplication, ie for everyone. But if so, one can show that with the relative weak -* topology is a locally compact Hausdorff space. According foregoing

A continuous homomorphism with norm. is the algebra of continuous, complex-valued functions that vanish at infinity. This homomorphism is called Gelfand transformation, called the Gelfand transform of a

Example C0 ( Z)

Let Z be a locally compact Hausdorff space and, then A is already an algebra of the type to which maps the Gelfand transform. To determine the Gelfand transform for this case, we need to gain an overview of the multiplicative functionals on A. If so, the point evaluation is obviously a multiplicative functional, and one can show that these are all already, ie that applies. Z can therefore be identified by means of the figure, at least as a quantity. One can show that this map is even a homeomorphism, so that you can identify Z as well as topological spaces. In this case, is nothing more than the identity. For the Gelfand transform offers nothing new.

Example L1 ( R)

The Banach space is the convolution as multiplication and the 1- norm a commutative Banach algebra -. For is paid

What are the multiplicative functionals on A? The point evaluations of the sample's out of the question, because for functions, the function value is not defined in one place. It can be shown that for by

A multiplicative functional on declared, and that, conversely, each of the multiplicative functional form. It is therefore necessary and one can further show that the mapping is a homeomorphism from to. If we identify and, therefore, by means of this figure, the Gelfand transform has the form:

The Gelfand transform proves to be an abstraction of the Fourier transform.

Example ' holomorphic continuation '

Let Z be the circle. Then a commutative Banach algebra with 1 Let A is the disc algebra, ie the subalgebra of all functions that have a holomorphic continuation to the interior. With a little function theory ( maximum principle ), one can show that A is a sub - Banach algebra of. What are the multiplicative functionals on A? First and multiplicative functionals on A. are the point evaluations that are already multiplicative functionals on, of course, but there are more. Since the holomorphic continuation of a function into the interior is unique, all point evaluations, multiplicative functionals on A. It is shown that and that can be identified topologically with the circular area means. In this example, therefore,

That is, the Gelfand transform plays the role of a continuation operator here.

Importance

If A is a commutative C * - algebra, then the Gelfand transform of the isometric isomorphism from the set of Gelfand - Neumark for commutative C * - algebras. This is the starting point of the spectral theory.

The example generalizes to locally compact abelian groups G. The Gelfand space of is denoted by and can be re- equipped with a group structure. It then calls the dual group of G. This is a starting point of abstract harmonic analysis.

The Stone - Čech compactification of a completely regular Hausdorff space X can be obtained as an application of the Gelfand transform on the commutative C * - algebra of continuous and bounded functions on X.

The core of the Gelfand transformation in the case of a commutative Banach the Jacobson - radical, in particular the Jacobson radical is always completed. This shows again, interlocking like algebraic and topological concepts in the theory of Banach algebras.