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Let $H_i, K_i$ for $i=1,2,3$ be Hilbert spaces with horizontal exact sequences. Assume $T_1, T_2, T_3$ have dense range, that $T_1, T_2, T_3$ are trace class operators and that the squares commute. Assume that $S \colon H_2 \to K_2$ is a compact operator with dense range making the squares commute.

$\require{AMScd}$ \begin{CD} 0 @>>> H_1 @>f_1>> H_2 @>f_2>> H_3 @>>> 0\\ @V VV @V T_1 VV @V T_2 VV @V T_3 VV @V VV \\ 0 @>>> K_1 @>>g_1> K_2 @>>g_2> K_3 @>>> 0 \end{CD}

Is it the case that there is a $C > 0$ such that $C \mathrm{Tr} |T_2| \geq \mathrm{Tr} |S|$?

This is related to the question: Are nuclear operators closed under extensions?

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  • $\begingroup$ What are your morphisms between Hilbert spaces? $\endgroup$ Commented May 31, 2020 at 14:17
  • $\begingroup$ Bounded operators $\endgroup$ Commented May 31, 2020 at 14:27

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The answer is no. Take $H_2 = K_2$ to be an infinite dimensional Hilbert space and take $H_1 = K_1$ to be an infinite dimensional subspace whose orthocomplement $H_3 = K_3$ is also infinite dimensional. Let $f_1 = g_1$ be the inclusion and $f_2 = g_2$ the orthogonal projection. Let $T_1: H_1 \to K_1$ and $T_3: H_3 \to K_3$ be any trace class operators with dense range such that ${\rm ker}(T_3)$ is infinite dimensional. Let $T_2 = T_1 \oplus T_3$.

Let $S_0: H_2 \to K_2$ be any operator which vanishes on ${\rm ker}(T_3)^\perp$ and takes ${\rm ker}(T_3)$ into $K_1$. Then $S = S_0 + T_2$ will make the diagram (with $T_1$ and $T_3$ unaltered, I am sure you meant) commute. So clearly the trace of $|S|$ can be arbitrarily large. Also, ${\rm ran}(S)$ contains both ${\rm ran}(T_1)$ and ${\rm ran}(T_3|_{{\rm ker}(T_3)^\perp}) = {\rm ran}(T_3)$, so the range of $S$ is dense.

(In the first version of this answer I didn't notice that the ranges had to be dense.)

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