# 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 $\ker(f_2)$"? May 24 '20 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. May 24 '20 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 May 24 '20 at 17:41

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. May 24 '20 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 May 24 '20 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)! May 24 '20 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. May 24 '20 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. May 26 '20 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 May 24 '20 at 17:47
• It also took me far too long... May 24 '20 at 17:51