It is well known in narrow circles that the homological dimension (in the sense of relative Banach homology) of $c_0$-module $\ell_\infty$ is 2. As the corollary, this module is not projective. This proof is rather involved, its main ingridient is a lack of a right inverse for the mapping: $$ \Delta:c_0\;\hat{\otimes}\;c_0\to(c\;\hat{\otimes}\;c_0)\oplus(c_0\;\hat{\otimes}\;\ell_\infty): x\;\hat{\otimes}\;y\mapsto (x\;\hat{\otimes}\;y)\oplus(x\;\hat{\otimes}\;y) $$ in the category of left Banach $c_0$-modules.

I would like to see a more direct proof of non-projectivtity. The standard route would be to show that there is no right inverse $c_0$-morphism for the mapping $\pi:c \;\hat{\otimes}\; \ell_\infty\to \ell_\infty \colon a\; \hat{\otimes}\; x\mapsto a\cdot x$, where $c$ is the Banach space of convergent sequences.

Does anyone have an idea how to prove non-projectivity more or less directly?

  • 1
    $\begingroup$ Just to make the question self-contained... How does $c_0$ act on $c\widehat\otimes \ell^\infty$? $\endgroup$ – Matthew Daws Oct 15 '18 at 14:50
  • $\begingroup$ @MatthewDaws, simply the elementwise multiplication. $\endgroup$ – Norbert Oct 15 '18 at 16:04
  • $\begingroup$ Yes, but on which tensor factor? $\endgroup$ – Matthew Daws Oct 15 '18 at 16:11
  • $\begingroup$ @MatthewDaws, on the first one by the formula $a(b\;\otimes\; x)=ab\;\otimes\; x$, where $a\in c_0, b\in c, x\in\ell_\infty$. $\endgroup$ – Norbert Oct 15 '18 at 17:27

I believe the following works...

Notice that $c=c_0 + \mathbb C1$ and so $$\newcommand{\proten}{\widehat\otimes} c\proten\ell^\infty = c_0\proten\ell^\infty + 1 \otimes \ell^\infty.$$ This is an isomorphism, maybe not isometric. Suppose, towards a contradiction, that there is a right inverse $T:\ell^\infty \rightarrow c\proten\ell^\infty$, so $T$ factors as $$ T(x) = T_1(x) + 1\otimes T_2(x)\qquad(x\in\ell^\infty), $$ where $T_1:\ell^\infty\rightarrow c_0\proten\ell^\infty$ and $T_2:\ell^\infty\rightarrow\ell^\infty$. That $T$ is a left $c_0$-module homomorphism means that $$ T(ax) = T_1(ax) + 1\otimes T_2(ax) = a\cdot T(x) =a\cdot T_1(x) + a\otimes T_2(x) \qquad (a\in c_0, x\in\ell^\infty). $$ Thus $T_2(a)=0$ for each $a\in c_0$ and $T_1(ax) = a\cdot T_1(x) + a\otimes T_2(x)$ for $a\in c_0, x\in\ell^\infty$. Finally, we should have that $\pi T(x)=x$, that is, $$ \pi_1 T_1(x) + T_2(x) = x \qquad (x\in\ell^\infty), $$ where $\pi_1:c_0\proten\ell^\infty\rightarrow\ell^\infty$ is the multiplication. Notice that $\pi_1$ takes value in $c_0$.

Then, for $a\in c_0$, as $T_2(a)=0$, we see that $\pi_1T_1(a)=a$. Thus $\pi_1T_1:\ell^\infty\rightarrow c_0\subseteq\ell^\infty$ is a projection, which is well-known not to exist. (This is Phillip's Lemma.)

| cite | improve this answer | |
  • $\begingroup$ Thank you for your nice solution and my apologies for delayed reply. I looked through your solution and realized that it could be generalized to the following proposition. If a Banach $A$-module $X$ is projective, then $\operatorname{span}(A X)$ is complemented in $X$. $\endgroup$ – Norbert Oct 27 '18 at 1:07

If $\ell^\infty$ was projective then any Banach limit $L:\ell^\infty \to \mathbb{C}$ would extend to a $c_0$-module morphism $\bar{L}:\ell^\infty \to c$ such that $L=\lim\circ \bar{L}$, but such does not exist.

Assume such $\bar{L}$ does exist. Consider $r=\bar{L}(1)-1\in c$ and observe that $\lim r=\lim\bar{L}(1)-1=L(1)-1=0$, thus $r\in c_0$. Let $I\subset \mathbb{N}$ be the set of indices $n$ for which $r_n=-1$ and let $V<\ell^\infty$ be the vector space of sequences supported on $I$. Note that $I$ is finite, thus $V$ is finite dimensional. For $s\in c_0$, $\bar{L}(s)=s\bar{L}(1)=sr+s$. It follows that $c_0\cap \ker\bar{L}=V$. But, as $V$ and $c$ are separable and $\ell^\infty$ is not, we can find $x\in \ker \bar{L}$ which is not in $V$. We consider the element $s=(1/n)\in c_0$ and observe that $sx$ is in $c_0\cap\ker\bar{L}$ but not in $V$. This is a contradiction.

| cite | improve this answer | |
  • $\begingroup$ This a good proof from the first principles. Thank you. $\endgroup$ – Norbert Oct 26 '18 at 16:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.