Gelfand transformation

The Gelfand transformation (after Israel Gelfand ) is the most important instrument in the theory of commutative Banach algebras. It maps a commutative - Banach algebra A into an algebra of continuous functions. A continuous function is assigned to each of them , whereby a suitable locally compact Hausdorff space is. The assignment is a continuous algebra homomorphism . ${\displaystyle \mathbb {C} }$ ${\displaystyle a}$ ${\displaystyle A}$ ${\displaystyle {\hat {a}}\colon X\to {\mathbb {C} }}$ ${\displaystyle X}$ ${\displaystyle a\mapsto {\hat {a}}}$

Motivation, Gelfand space

If one considers a commutative - Banach algebra only as a standardized space with dual space and bidual space , then the elements of can be mapped to continuous functions as follows: Assign each function to. This is the well-known isometric embedding of in the dual space, because each is an element from . Each is also steady. The standard topology proves to be unnecessarily strong. For this reason one looks at the weak - * - topology , this is just defined as the coarsest topology that makes all mappings continuous. ${\displaystyle \mathbb {C} }$ ${\displaystyle A}$ ${\displaystyle A'}$ ${\displaystyle A''=(A')'}$ ${\displaystyle A}$ ${\displaystyle a}$ ${\displaystyle {\hat {a}}\colon A'\rightarrow {\mathbb {C} },{\hat {a}}(\varphi ):=\varphi (a)}$ ${\displaystyle A}$ ${\displaystyle {\hat {a}}}$ ${\displaystyle A''}$ ${\displaystyle {\hat {a}}}$ ${\displaystyle A'}$ ${\displaystyle {\hat {a}}}$

If we turn back to algebra , we must state that the assignment is not a homomorphism ; it is not multiplicative, i.e. H. it does not apply . For this would have to apply to everyone , but a linear functional is usually not multiplicative. However, this observation gives an indication of how one can construct a homomorphism of the desired kind. Instead of using only the multiplicative functionals in , and that is exactly what the Gelfand transformation is. ${\displaystyle A}$ ${\displaystyle a\mapsto {\hat {a}}}$ ${\displaystyle {\widehat {ab}}={\hat {a}}{\hat {b}}}$ ${\displaystyle {\widehat {ab}}(\varphi )={\hat {a}}(\varphi ){\hat {b}}(\varphi )}$ ${\displaystyle \varphi (ab)=\varphi (a)\varphi (b)}$ ${\displaystyle \varphi \in A'}$ ${\displaystyle A'}$ ${\displaystyle A'}$

We therefore set . This amount is called the spectrum ( Gelfand spectrum ) of or also the Gelfand space of . Note that the zero homomorphism has been removed. There are Banach algebras with an empty spectrum, e.g. B. a Banach algebra with zero multiplication, d. H. for everyone . If, however , one can show that with the relative weak - * - topology there is a locally compact Hausdorff space . According to the above is ${\displaystyle X_{A}:=\{\varphi \in A';\varphi \,{\text{multiplikativ}},\varphi \not =0\}}$ ${\displaystyle A}$ ${\displaystyle A}$ ${\displaystyle A}$ ${\displaystyle a\cdot b=0}$ ${\displaystyle a,b\in A}$ ${\displaystyle X_{A}\not =\emptyset }$ ${\displaystyle X_{A}}$

${\displaystyle {\mathcal {G}}:A\rightarrow C_{0}(X_{A}),\,a\mapsto {\hat {a}},\,{\hat {a}}(\varphi )=\varphi (a)}$

a continuous homomorphism with norm . is the algebra of continuous, complex-valued functions that vanish at infinity . This homomorphism is called the Gelfand transform, it is called the Gelfand transform of . ${\displaystyle \leq 1}$ ${\displaystyle C_{0}(X_{A})}$ ${\displaystyle X_{A}}$ ${\displaystyle {\hat {a}}}$ ${\displaystyle a}$

Example C ₀ (Z)

Let Z be a locally compact Hausdorff space and then A is already an algebra of the kind to which the Gelfand transformation maps. In order to determine the Gelfand transformation for this case, we have to get an overview of the multiplicative functionals on . If , then the scoring is evidently a multiplicative functional, and it can be shown that all of these are already there; that is , that applies. Z can by means of the image so to be identified, at least as a quantity. One can show that this mapping is even a homeomorphism , so that Z and can also be identified as topological spaces . So in this case it is nothing more than identity. The Gelfand transformation offers nothing new for. ${\displaystyle A=C_{0}(Z)}$ ${\displaystyle A}$ ${\displaystyle z\in Z}$ ${\displaystyle \delta _{z}:A\rightarrow {\mathbb {C} },\,\delta _{z}(f):=f(z)}$ ${\displaystyle X_{A}=\{\delta _{z};z\in Z\}}$ ${\displaystyle z\mapsto \delta _{z}}$ ${\displaystyle X_{A}}$ ${\displaystyle X_{A}}$ ${\displaystyle {\mathcal {G}}:A\rightarrow C_{0}(X_{A})=C_{0}(Z)}$ ${\displaystyle A=C_{0}(Z)}$

Example L ¹ (ℝ)

With the convolution as multiplication and the 1-norm, the Banach space is a commutative -Banach algebra. For is paid ${\displaystyle A=L^{1}(\mathbb {R} )}$ ${\displaystyle \mathbb {C} }$ ${\displaystyle f,g\in L^{1}(\mathbb {R} )}$

${\displaystyle f*g(t):=\int _{-\infty }^{\infty }f(s)g(t-s)\mathrm {d} s}$

${\displaystyle \|f\|_{1}=\int _{-\infty }^{\infty }|f(s)|\mathrm {d} s}$

What do the multiplicative functionals look like ? The point evaluations in the example are out of the question, because the function value is not defined at one point for functions. You can show that for by ${\displaystyle A}$ ${\displaystyle C_{0}(Z)}$ ${\displaystyle L^{1}}$ ${\displaystyle z\in {\mathbb {R} }}$

${\displaystyle \varphi _{z}(f):=\int _{-\infty }^{\infty }f(t)e^{-itz}\mathrm {d} t}$

a multiplicative functional is declared on , and that conversely every multiplicative functional is of this form. So it is true and one can further show that the mapping is a homeomorphism of on . If one identifies therefore and by means of this mapping, the Gelfand transformation has the form: ${\displaystyle A=L^{1}(\mathbb {R} )}$ ${\displaystyle X_{A}=\{\varphi _{z};z\in \mathbb {R} \}}$ ${\displaystyle z\mapsto \varphi _{z}}$ ${\displaystyle \mathbb {R} }$ ${\displaystyle X_{A}}$ ${\displaystyle {\mathbb {R} }}$ ${\displaystyle X_{A}}$

${\displaystyle {\mathcal {G}}:L^{1}(\mathbb {R} )\rightarrow C_{0}(\mathbb {R} ),\,f\mapsto {\hat {f}},\,{\hat {f}}(z)=\int _{-\infty }^{\infty }f(t)e^{-itz}\mathrm {d} t}$.

The Gelfand transformation thus proves to be an abstraction of the Fourier transformation .

Example 'holomorphic continuation'

It is the circle line . Then there is a commutative Banach algebra with 1. Let the disk algebra , that is, the subalgebra of all functions that have a holomorphic continuation into the interior . With a little function theory ( maximum principle ) one shows that is a sub-Banach algebra of . What do the multiplicative functionals look like ? First of all, the point evaluations , which are already multiplicative functionals open , are of course also multiplicative functionals open . But there are others. Since the holomorphic continuation of a function inside is unambiguous, all point evaluations , multiplicative functionals, are on . One shows that and that one can also identify topologically with the circular area using . In this example, therefore ${\displaystyle Z}$ ${\displaystyle \{z\in \mathbb {C} ;|z|=1\}}$ ${\displaystyle C_{0}(Z)}$ ${\displaystyle A}$ ${\displaystyle \{z\in \mathbb {C} ;|z|<1\}}$ ${\displaystyle A}$ ${\displaystyle C_{0}(Z)}$ ${\displaystyle A}$ ${\displaystyle \delta _{z},|z|=1}$ ${\displaystyle C_{0}(Z)}$ ${\displaystyle A}$ ${\displaystyle \delta _{z},|z|<1}$ ${\displaystyle A}$ ${\displaystyle X_{A}=\{\delta _{z};|z|\leq 1\}}$ ${\displaystyle X_{A}}$ ${\displaystyle z\mapsto \delta _{z}}$ ${\displaystyle K=\{z\in \mathbb {C} ;|z|\leq 1\}}$

${\displaystyle {\mathcal {G}}\colon A\rightarrow C_{0}(K),\,f\mapsto {\hat {f}},\,{\hat {f}}=}$holomorphic continuation of , ${\displaystyle f}$

d. H. the Gelfand transformation plays the role of a continuation operator here.

meaning

If a commutative C * -algebra , then the Gelfand transformation is the isometric isomorphism from the Gelfand-Neumark theorem for commutative C * -algebras. That is the starting point of spectral theory . ${\displaystyle A}$

The example generalizes to locally compact, Abelian groups . The Gelfand room of is labeled with and can be given a group structure again. One then calls the dual group of . This is a starting point for abstract harmonic analysis . ${\displaystyle L^{1}({\mathbb {R} })}$ ${\displaystyle G}$ ${\displaystyle L^{1}(G)}$ ${\displaystyle {\hat {G}}}$ ${\displaystyle {\hat {G}}}$ ${\displaystyle G}$

The Stone-Čech compactification of a completely regular Hausdorff space can be obtained as an application of the Gelfand transformation to the commutative C * algebra of the continuous and bounded functions on . ${\displaystyle X}$ ${\displaystyle C_{b}(X)}$ ${\displaystyle X}$

In the case of a commutative Banach algebra, the core of the Gelfand transformation is the Jacobson radical , in particular the Jacobson radical is always closed. This shows again how algebraic and topological terms interlock in the theory of Banach algebras.

literature

• RV Kadison , JR Ringrose : Fundamentals of the Theory of Operator Algebras , 1983
• M. Takesaki : Theory of Operator Algebras I (Springer 1979, 2002)