# Explicit proof that $c_0$-module $\ell_\infty$ is not projective

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?

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

• 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 $X$. 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.

• This a good proof from the first principles. Thank you. Oct 26 '18 at 16:01