# Are extensions of nuclear Fréchet spaces nuclear?

Consider the category of Fréchet spaces, the morphisms being continuous linear maps with closed image. Suppose that we have a short exact sequence in that category:

$0 \rightarrow V_1 \rightarrow V_2 \rightarrow V_3 \rightarrow 0$.

Of course $V_1$ and $V_3$ are nuclear if $V_2$ is. I recently asked myself if the converse might be true. I haven't found anything useful in the standard literature (Treves, Schaefer) but that might be just me being too ignorant to see the obvious. I'm grateful if someone could shed some light on this.

Cheers,

Ralf

• I suspect that nuclearity of $V_3$ might force the extension to split, because nuclear Fr\'echet spaces have some kind of lifting property... does this sound like it might work? Mar 31, 2011 at 20:19
• Yemon, this isn't true. See my comment on Ralf's answer. Apr 1, 2011 at 7:34
• Your "category" isn't a category because morphisms with closed image aren't closed under composition. Indeed, every continuous linear map $f: X \to Y$ is the composition of $X \to X \times Y, x \mapsto (x,f(x))$ which has closed image because the graph is closed by continuity and the projection $X \times Y \to Y$ which also has closed image. You can still speak of short exact sequences by requiring that the left morphism be the kernel of the right morphism and that the right morphism be the cokernel of the left morphism. Google for quasi-abelian categories and exact categories for more on this Feb 2, 2012 at 17:21
• Theo, is the cokernel in the category of Frechet spaces and continuous linear maps really what we want in this case? Don't we want a smaller class of "exact sequences", namely those which are exact as diagrams in Vect? (I guess this is addressed in eg your Expositiones article) Feb 2, 2012 at 18:10

You question was answered even for locally convex spaces by S. Dierolf and W. Roelcke in proposition 3.8 of the article "On the three-space-problem for topological vector spaces". Collect. Math. 32, p. 13-35 (1981).

The splitting theory for Frechet spaces is nowadays very well understood by results of D. Vogt and others. This can be found in my "Derived functors in functional analysis", Springer Lecture Notes in Mathematics 1810 (2003).

• But answered which way? Could you add a brief note at least saying what the answer actually is so that those casually reading this, or those without access to the references, can at least know what the answer is. Feb 1, 2012 at 9:28
• In Ralf's question, $V_2$ is nuclear whenever $V_1$ and $V_3$ are nuclear. This can be reformulated without exact sequences: A locally convex space is nuclear whenever it has a nuclear subspaces such that the corresponding quotient is also nuclear. In Dierolf's and Roelcke's article there many more results of this type as well as many counterexamples. Feb 1, 2012 at 9:40

Thank you for the hint, Yemon. You are indeed right. I found the following paper which proves the lifting property that you mentioned: emis.de/journals/PM/55f1/pm55f107.ps.gz The splitting follows from exmaple 3 on p. 96. Thanks again :-)

• Glad to hear that my vague memory was correct - I wrote the comment not to be cryptic, but because I was in a rush earlier and didn't have time to chase down the references. Mar 31, 2011 at 22:06
• Example 3 on p96 says "Each nuclear DF -space has the lift property for the class of Frechet spaces" so it **does not apply. Indeed, here's an example of a short exact sequence of nuclear Frechet spaces that does not split: $L_\flat \mathbb{R} \to L\mathbb{R} \to \mathbb{R}^{\mathbb{N}}$. The middle is smooth loops in R and the left-hand is smooth loops that are infinitely flat at the identity. This does not split, but all are nuclear Frechet spaces. Apr 1, 2011 at 7:34
• Indeed, I read in the introduction to that paper "Gejler has proved that a nuclear Frechet space has the lifting property for the class of all nuclear Frechet spaces if and only if it is finite dimensional." Apr 1, 2011 at 8:00
• Oh, my bad and a prime example of wishful reading. Thank you for the correction and the counterexample, Andrew.
– Ralf
Apr 13, 2011 at 0:37