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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.

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    $\begingroup$ Just to get the terminology straight in my head: by topologically short exact do you mean the morphisms are bounded linear maps but you have short exactness in the category of vector spaces and linear maps? In other words, $f_1$ has closed range, and not just "$f_1(X_1)$ is dense in $\ker(f_2)$"? $\endgroup$ – Yemon Choi May 24 at 16:51
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    $\begingroup$ I mean the first, i.e. $f_1$ has closed range. In that sense what I mean by "topological exactness" is algebraic exactness. $\endgroup$ – santker heboln May 24 at 17:09
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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.

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  • $\begingroup$ That's the case $A=0$ of my example, I believe :) But yours is a cleaner and simpler approach $\endgroup$ – Yemon Choi May 24 at 17:41
  • $\begingroup$ Your answer appeared while I was writing mine. I had to look for the latex code. $\endgroup$ – M.González May 24 at 17:44
  • $\begingroup$ Sure, no worries. I think several of us had the same idea independently and simultaneously $\endgroup$ – Yemon Choi May 24 at 17:45
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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.


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 $\theta=T_2$.

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  • $\begingroup$ This was a quick response! In fact I just wanted to edit my question, to say that $T_1, T_2$ and $T_3$ should be surjective. Your answer is interesting nonetheless. $\endgroup$ – santker heboln May 24 at 17:37
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    $\begingroup$ @santkerheboln before you edit the question, I should point out that surjective operators on Banach spaces are nuclear if and only if they are finite-rank. So your modified question will have a positive answer for trivial reasons $\endgroup$ – Yemon Choi May 24 at 17:43
  • $\begingroup$ Indeed, I guess a more sensible assumption (to get a positive result and also in view of intended applications) would be that the squares commute. In any case, the question as stated has been answered exhaustively. Thank you (all)! $\endgroup$ – santker heboln May 24 at 17:59
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    $\begingroup$ I now think the conditions the $T_i$ having dense ranges for $i = 1,2,3$ and the squares commuting should give the desired equality $\theta = T_2$ in your approach. $\endgroup$ – santker heboln May 24 at 22:35
  • $\begingroup$ Addendum: Actually, Jochen's example shows that even if $T_i$'s are assumed to be surjective, $\theta$ may not equal $T_2$. However, if $T_i$ have dense ranges, I think $\theta$ could be "close" to $T_2$, it seems unitarily equivalent. $\endgroup$ – santker heboln May 26 at 2:07
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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 $b$.

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  • $\begingroup$ Actually I spent an embarrassingly long time trying to turn the observations mentioned at the end of my answer into a proof that the original question had a positive answer. It wasn't until I tried to work out why I was getting stuck that the basic point observed by you and M.Gonzalez hit me $\endgroup$ – Yemon Choi May 24 at 17:47
  • $\begingroup$ It also took me far too long... $\endgroup$ – Jochen Wengenroth May 24 at 17:51

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