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hom-functor [2014/06/23 22:57] nikolaj |
hom-functor [2014/11/01 16:48] nikolaj |
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Since locally compact groups have the Haar measure, one can form the space of integrable functions $G\to\mathbb C$ and carry it along with these functors. This procedure is the Fourier transform and it really works for all groups in this category! E.g. for $U(1)$ itself, i.e. the periodic interval, we find that the dual group is $\mathbb Z$ and the associated transform is the Fourier series. | Since locally compact groups have the Haar measure, one can form the space of integrable functions $G\to\mathbb C$ and carry it along with these functors. This procedure is the Fourier transform and it really works for all groups in this category! E.g. for $U(1)$ itself, i.e. the periodic interval, we find that the dual group is $\mathbb Z$ and the associated transform is the Fourier series. | ||
- | Extending this to a category of non-commutative groups is a research subject. Tannakian categories ... Grothendieck stuff ... we see that it's desirable for the Hom-sets to have their own algebraic structure (in the above, the Hom-set was again a group) and this is where many ideas come from. | + | == More general cases == |
+ | - Extending this to a category of non-commutative groups is a research subject. Tannakian categories ... Grothendieck stuff ... we see that it's desirable for the Hom-sets to have their own algebraic structure (in the above, the Hom-set was again a group) and this is where many ideas come from. | ||
+ | - A similar construction works if we consider commutative Banach algebra and their maps maps to $\mathbb C$. This is the Gelfand transform business. It reduces to the Fourier transform if we consider the space $L^1(\mathbb R)$ and convolution as multiplication. | ||
=== Yoneda === | === Yoneda === | ||
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simply because their values lie in the field $\mathbb K$, which already has addition. And as a side note, if we restrict ourselves to the linear functionals, then this functional space becomes the so called dual vector space. | simply because their values lie in the field $\mathbb K$, which already has addition. And as a side note, if we restrict ourselves to the linear functionals, then this functional space becomes the so called dual vector space. | ||
- | Now back to the general case. Instead of working with the category ${\bf C}$, one can work with the category of set valued covariant functors ${\bf Set}^{{\bf C}^\mathrm{op}}$ called [[Presheaf category]]). One substitutes an object $A$ with the contravariant functor $\mathrm{Hom}_{\bf C}(-,A)$ (=[[Yoneda embedding]]) and the arrows actually are the same/isomorphic to the old ones (=Yoneda lemma). | + | Now back to the general case. Instead of working with the category ${\bf C}$, one can work with the category of set valued covariant functors ${\bf Set}^{{\bf C}^\mathrm{op}}$ (called [[Presheaf category]]). One substitutes an object $A$ with the contravariant functor $\mathrm{Hom}_{\bf C}(-,A)$ (=[[Yoneda embedding]]) and the arrows actually are the same/isomorphic to the old ones (=Yoneda lemma). |
The advantage of this is that that new category ${\bf Set}^{{\bf C}^\mathrm{op}}$ has more objects than ${\bf C}$, namely some non-representable functors $F$. (Representable means $F$ is isomorphic to some hom-functor anyway.) For example, the category ${\bf C}$ might not have products $\times$, but because ${\bf Set}$ has products, the category ${\bf Set}^{{\bf C}^\mathrm{op}}$ always has them. | The advantage of this is that that new category ${\bf Set}^{{\bf C}^\mathrm{op}}$ has more objects than ${\bf C}$, namely some non-representable functors $F$. (Representable means $F$ is isomorphic to some hom-functor anyway.) For example, the category ${\bf C}$ might not have products $\times$, but because ${\bf Set}$ has products, the category ${\bf Set}^{{\bf C}^\mathrm{op}}$ always has them. |