Are nuclear operators closed under extensions?

Question

Given $X_i, Y_i$ Banach spaces, $f_j, g_j, T_i$ bounded linear operators for $i=1,2,3$ and $j=1,2$. We have the following diagram

$\require{AMScd}$ \begin{CD} 0 @>>> X_1 @>f_1>> X_2 @>f_2>> X_3 @>>> 0\\ @V VV @V T_1 VV @V T_2 VV @V T_3 VV @V VV \\ 0 @>>> Y_1 @>>g_1> Y_2 @>>g_2> Y_3 @>>> 0 \end{CD}

with two horizontal topologically short exact sequences. If $T_1$ and $T_3$ are nuclear operators, does it imply that $T_2$ is nuclear as well? A reference to problems of this general form, would be most welcome.

Answer 1

The answer is no: you can even have $T_1=T_3=0$ and $T_2$ equal to the identity $id$ on an infinite dimensional Banach space.

Indeed, consider the following commutative diagram with exact rows:

$$\begin{CD} 0@>>> 0 @>0>> X @>id>> X @>>> 0\\ &&@V0VV @VV{id}V @VV0V\\ 0@>>>X @>>id> X @>>0> 0 @>>> 0 \end{CD} $$

See this paper for related results.

Answer 2

In the generality stated, the question has a negative answer (which took me embarrassingly long to spot); the point is that when $T_1$ and $T_3$ are not assumed injective or surjective they give us very little traction on what $T_2$ is, in contrast to the intuition one might have from the Five Lemma. The "silly" counterexample might be useful for other settings so here are the details; it should work in any category with zero object and an appropriate notion of "binary sum".

Take objects (=Banach spaces if you wish to be concrete) $A$ and $B$ and let $\iota_L: A\to A\oplus B$, $\pi_L: A\oplus B \to A$, $\iota_R:B\to A\oplus B$ and $\pi_R: A\oplus B \to B$ be the usual embedding and projection operators.

Take as your top row the short exact sequence $A \stackrel{\iota_L}{\to} A\oplus B \stackrel{\pi_R}{\to} B$ and as your bottom row the short exact sequence $B \stackrel{\iota_R}{\to} A\oplus B \stackrel{\pi_L}{\to} A$. The vertical arrow on the left is the zero map $A\to B$, the vertical arrow on the right is the zero map $B\to A$, and the middle arrow is $(0,{\rm id_B}) : A\oplus B \to A\oplus B$. Then everything commutes.

To turn this into a counterexample for the original question, just take $B$ to be your favourite infinite-dimensional Banach space and $A$ to be an arbitrary Banach space.

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On the other hand, I think I can prove the following: suppose I am given everything in your initial diagram except the middle arrow $T_2$, so that we have (strict) exactness on the top row and bottom row and nuclear operators $T_1:X_1\to Y_1$, $T_3:X_3\to Y_3$. Then $T_1$ has a nuclear extension $R:X_2\to Y_1$, $T_3$ has a nuclear lift $S:X_3\to Y_2$, and defining $\theta= g_1R+Sf_2$ gives a "middle arrow" which is nuclear and does make all the squares commute. So depending on your intended applications, this might be of some use; it says we can manufacture an "extension" of $T_1$ and $T_3$ which is nuclear. Furthermore, even if one wants to show that a given $T_2$ is nuclear, this construction might help since under certain extra conditions one may be able to prove that $\theta=T_2$.

Answer 3

Of course, Yemon was faster than me. But I want to emphasize that the point is very elementary linear algebra: The simple commutative diagram

$\begin{CD} 0 @>>> \mathbb R @>f_1>> \mathbb R^2 @>f_2>> \mathbb R @>>> 0\\ @V VV @V T_1 VV @V T_2 VV @V T_3 VV @V VV \\ 0 @>>> \mathbb R @>>f_1> \mathbb R^2 @>>f_2> \mathbb R @>>> 0 \end{CD}$

with the natural inclusion $f_1(x)=(x,0)$ and projection $f_2(x,y)=y$ in the rows shows shows that $T_2$ is not at all determined by $T_1$ and $T_3$. If $T_2$ is given by a matrix $\begin{bmatrix} a&b\\ 0&c\end{bmatrix}$ you get nothing for $b$.