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Is there a non trivial sequence $(T_{n})$ of linear operators $T_{n}$ on $M_{n}(\mathbb{C})$ such that $$T_{nm}(X\otimes Y)=T_{n}(X) \otimes T_{m}(Y) $$ where $X$ and $Y$ are in $M_{n}(\mathbb{C})$ and $M_{m}(\mathbb{C})$, respectively?

By trivial sequence we mean $T_{n}=Id$ for all $n$.

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3 Answers 3

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$T_n(X) = n \times n$ matrix with all entries $trace(X)$.
This doesn't satisfy $T_n(X_1 X_2) = T_n(X_1) T_n(X_2)$.

Edit: More examples:

  1. $T_n(X)$ has $trace(X)$ in upper left corner, 0 elsewhere.

  2. $T_n(X)$ has $trace(X)$ in lower right corner, 0 elsewhere.

  3. $T_n(X)$ has sum of anti-diagonal $(\Sigma x_{i,n-i})$ in upper right corner, 0 elsewhere.

  4. $T_n(X)$ has sum of anti-diagonal in lower left corner, 0 elsewhere.
  5. $T_n(X)$ has all entries = sum of all entries of $X$.

Edit: Another type:

  1. $T_n(X)=0$ ($n$ even); $T_n(X)$ has sum of middle row in every middle row entry, 0 elsewhere ($n$ odd).

  2. Same sum as in (6.) except $T_n(X)=0$ has only one non-zero entry at left end resp. middle resp. right end of middle row.

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  • $\begingroup$ [This doesn't satisfy $T_n(X_1 X_2) = T_n(X_1) T_n(X_2)$.]---> Yes but this is not required and your example fits perfectly the requirements of the MO (+1). We can even mix your $T_n$ with thee ``logarithmic trick'' of Simon. $\endgroup$
    – Duchamp Gérard H. E.
    Commented Apr 26, 2016 at 11:46
  • $\begingroup$ [We can even mix...] ---> Yes: if $(S_n), (T_n)$ work then so does $(S_{n}T_{n})$. This example and the "logarithmic trick" examples all commute. $\endgroup$
    – David Lampert
    Commented Apr 26, 2016 at 13:04
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What is a linear operator---do you require $T_n(X_1 X_2) = T_n(X_1) T_n(X_2)$? If not, $T_n(X) = c^n X$ for a fixed scalar $c$ works. Even if yes, I think that $T_n(X) = X$ if $n$ odd, $0$ if $n$ even should work ($T_n = Id$ for odd $n$, but not all $n$).

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  • $\begingroup$ Thanks for your answer. Your last example works. but the first one does not work, for n=1 unless c=1. $\endgroup$
    – Ali Taghavi
    Commented Apr 26, 2016 at 8:24
  • $\begingroup$ A revised version : is every sequence necessarily a sequence of 0 ,Id? $\endgroup$
    – Ali Taghavi
    Commented Apr 26, 2016 at 8:31
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Pick for each prime number $p$ a complex number $\alpha_p$.

For $n$ an integer, define $\alpha_n = \prod \alpha_p ^{\nu_p(n)}$ where $\nu_p(n)$ is the order of $p$ in the prime decomposition of $n$.

One has $\alpha_{nm} = \alpha_n \alpha_m$

Then you can take $T_n(M) = \alpha_n M$

Zach Teitler's example correspond to $\alpha_2=0$ and $\alpha_p =1$ for all odd prime $p$.

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  • $\begingroup$ Thanks for your very interesting answer. what about if a trivial sequence would be meant as "a sequence of scalar operator"? is there a non trivial example? $\endgroup$
    – Ali Taghavi
    Commented Apr 26, 2016 at 9:02
  • $\begingroup$ Maybe you can do the same construction with $\alpha_p$ an endomorphism of $M_p(\mathbb{C})$ for each prime $p$ ? But you need to be more carefull on how you identify $M_n \otimes M_m$ with $M_{nm}$ to see if this works... I'm not convinced. $\endgroup$
    – Simon Henry
    Commented Apr 26, 2016 at 10:18

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