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Injective Banach spaces, with morphisms as contractive linear maps, have been classically studied (and are $C(K)$ spaces with $K$ Stonian). But what about projectives?

So $P$ will be projective if given Banach spaces $E$ and $F$ and a quotient map (aka metric surjection) $\psi:E\rightarrow F$, given any contractive $\phi:P\rightarrow F$, we can lift this to a contractive $\varphi:P\rightarrow E$ with $\psi\varphi = \phi$. (If someone can make a nice commutative diagram, go ahead and edit this!)

Claim: The scalar field (say $\mathbb C$, but also works for $\mathbb R$) is not projective.

Proof: Let $f:c_0\rightarrow\mathbb C$ be the contractive functional $f(x) = \sum_{n=1}^\infty 2^{-n} x_n$ for $x=(x_n)\in c_0$. This induces an isometric isomorphism $c_0 / \ker(f) \cong \mathbb C$. So if $\mathbb C$ is projective, we can find a contractive $g:\mathbb C\rightarrow c_0$ with $fg=1$. That is, $g=(g_n)\in c_0$ is a norm-one vector with $f(g) = \sum 2^{-n} g_n =1$, which is impossible.

Question: Are there any projective Banach spaces?

It seems to me that the problem is insisting upon contractive morphisms. In the proof above, if we just need, for each $\epsilon>0$, to find $g$ with $\|g\|<1+\epsilon$, then this is no problem. Is anything known in this generalised setting? (It's easy to see that then $\ell_1(\Gamma)$ is always projective, if I allow myself this wiggle-room).

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See nlab.mathforge.org/nlab/show/projective+Banach+space which I had started in a momentary lapse of reason before sinking into the pit of teaching –  Yemon Choi Mar 30 '11 at 19:49
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The trick (if you want to get some notion of projectivity in the category I like to call Bang, as opposed to the category I like to call Bant) seems to be that one should ask not for lifting all metric surjections, but only to lift particular kinds of metric surjections; this is the approach in e.g. Semadeni's book, which is mentioned in the nLab entry linked to in my previous comment. –  Yemon Choi Mar 30 '11 at 19:53
    
There is an adjunction between Bang and Set where the forgetful functor is actually the unit ball functor; and this accounts for the fact that the admissible epimorphisms you wish to use in the definition of (relatively) projective have to be those maps from $X$ to $Y$ which induce surjections of the corresponding unit balls. –  Yemon Choi Mar 30 '11 at 20:03
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2 Answers

up vote 2 down vote accepted

You essentially answered your first question yourself: the ground field is a (contractive) retract of any nonzero Banach space by Hahn-Banach. If there were a non-zero projective Banach space in your sense then the ground field would be projective as well.

On the other hand: It is a theorem due to Köthe and Pełczyński that every projective Banach space (lifting over surjective maps in the additive category of Banach spaces) is isomorphic to $\ell^1{(S)}$ for some $S$. I don't know how well the norm of the isomorphism is controlled, but as all these spaces already satisfy your relaxed condition, you won't find any others.

The references for the second paragraph are:

  • A. Pełczyński, Projections in certain Banach spaces, Studia Math. 19 (1960), 209–228. MR0126145.
  • G. Köthe, Hebbare lokalkonvexe Räume, Math. Ann. 165 (1966), 181–195. MR0196464.
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By the way: My preferred way (of course very related to yours) of seeing that $\mathbb{R}$ is not contractively projective is to observe that the functional $\varphi: f \mapsto \int_{0}^{1/2} f - \int_{1/2}^{1} f$ on $C[0,1]$ has norm one, but the supremum in the definition of the norm is not a maximum. In other words, the identity of $\mathbb{R}$ does not lift to a contraction $\mathbb{R} \to C[0,1]$ over $\varphi$. –  Theo Buehler Mar 30 '11 at 11:41
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Comment for Matt: Regarding the K\"othe paper (wenn man kann nicht Deutsch sprechen) see also some of the answers and comments to mathoverflow.net/questions/5597 –  Yemon Choi Mar 30 '11 at 19:50
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Thanks, Yemon (why was I so sure that you would have something to say about this?!). If I remember correctly, Köthe's paper essentially proceeds by using a transfinite version of Pelczynski's marvellous rendition of the Eilenberg swindle, so to get the main idea it fully suffices to understand Pelczynski's paper. But it is a long time since I read both these papers, so I might misremember. Very minor nitpick: "Wenn man kein Deutsch spricht" sounds better to my Swiss ears ;) –  Theo Buehler Mar 30 '11 at 20:51
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Matthew, you answered your own question with your claim--there are no isometrically projective Banach space (since, for example, a contractively complemented subspace of a projective space is again projective).

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And with "no" you mean "no nontrivial". –  Martin Brandenburg Mar 30 '11 at 12:32
    
Picking up an old thread --- maybe no one is listening? A projective object $P$ in category theory is traditionally defined by the requirement that given any epimorphism $q:A\to B$ and any morphism $f:P\to B$ there exists a morphism $g:P\to A$ such that $f=qg$. In the categories of Banach spaces (with contractive operators for the "metric" category, and just bounded linear operators for the "isomorphic" category), the epimorphisms are just the morphisms with dense range. When we discuss projective Banach spaces, it seems we are not using the standard categorical definition of projectives. –  Fred Dashiell May 3 '13 at 12:05
    
As Bill said, there will be no projectives in the "metric" category of Banach spaces. But in addition, there will be no projectives in the isomorphic category either --- using the standard categorical sense of the term. This has always bothered me when we say that $\ell^1(\Gamma)$ is "projective". –  Fred Dashiell May 3 '13 at 12:10
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