# Are nuclear operators closed under extensions?

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.

• 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 $\ker(f_2)$"? – Yemon Choi May 24 at 16:51
• I mean the first, i.e. $f_1$ has closed range. In that sense what I mean by "topological exactness" is algebraic exactness. – santker heboln May 24 at 17:09

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.

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

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 $$\theta=T_2$$.

• 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. – santker heboln May 24 at 17:37
• @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 – Yemon Choi May 24 at 17:43
• 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)! – santker heboln May 24 at 17:59
• 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. – santker heboln May 24 at 22:35
• 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. – santker heboln May 26 at 2:07

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 $$b$$.

• 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 – Yemon Choi May 24 at 17:47
• It also took me far too long... – Jochen Wengenroth May 24 at 17:51