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Let $F$ - is functor in category of Banach spaces $\mathbf{Ban}$ with follow property: $f_n : A_n \to A_{n+1}$ exact sequence iff $F f_n : F A_n \to F A_{n+1}$ is exact sequence. Is it true, that $F$ naturally equivalent to functor of Banach adjoint $\mathbf{(\cdot)^*}$ or identity functor $\mathbf{1}$?

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  • $\begingroup$ What are the maps in this category? Bounded linear maps? Linear maps of norm $\leq 1$? $\endgroup$ – Eric Wofsey Nov 7 '15 at 17:37
  • $\begingroup$ @EricWofsey Bounded linear maps. $\endgroup$ – kp9r4d Nov 7 '15 at 17:40
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This is definitely not true. For instance, you could take $F(A)=A\oplus A$ (with any reasonable choice of norm). More generally, you should morally expect any functor of the form $F(A)=B\otimes A$ for some fixed nonzero $B$ to have this property, though I don't know enough about tensor products of infinite-dimensional Banach spaces to be able to say how much sense this makes if $B$ is infinite-dimensional.

(Also, the adjoint functor is not even a functor $\mathbf{Ban}\to\mathbf{Ban}$; it is a functor $\mathbf{Ban}\to\mathbf{Ban}^{op}$, i.e. a contravariant functor from $\mathbf{Ban}$ to itself.)

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  • $\begingroup$ Thanks! Did you know, there is exists some axiomatic characterization of Banach adjoint functor? For example: $F$ is Banach adjoint iff $F$ puts exact sequence into exact sequence and coretractions into retractions, (it is toy example, not serious proposition). $\endgroup$ – kp9r4d Nov 7 '15 at 17:59
  • $\begingroup$ Yes, of course you are right about contravariancy. $\endgroup$ – kp9r4d Nov 7 '15 at 18:08
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    $\begingroup$ I don't know of such a characterization off the top of my head. I would think that for any such characterization, one property you would need is that it sends $1$-dimensional spaces to $1$-dimensional spaces. I also think you would need some "infinitary" condition, something along the lines of "$F$ sends infinite direct sums to infinite direct products" (except that I don't know the right way to formulate this, since of course for Banach spaces there are many different flavors of infinite direct sums). $\endgroup$ – Eric Wofsey Nov 7 '15 at 18:08
  • $\begingroup$ Thanks a lot! I will think about your words. $\endgroup$ – kp9r4d Nov 7 '15 at 18:15

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