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What is an example of a coalgebra $C$ which admit a linear operator $T$, different from scalar operators $T=\lambda Id$, which satisfy $$(T\otimes T)\circ \Delta=\Delta \circ T^2 $$ but $C$ is not isomorphic to a subcoalgebra of $\mathbb{C}[x]$ or the trigonometric coalgebra generated by $c=cos, s=sin$?

Note that the differentiation operator on these coalgebras satisfies the above functional equation.

Our next question is the following: Does the above functional equation implies the automatic continuity of $T$ when $T$ is a symmetric operator on a quantum group?(By symmetricity of $T$ we mean $T(a^*)={(T(a))}^*$

Edit: According to the comment of Konstantinos, I add this link as a motivation for this question:

http://bims.iranjournals.ir/article_872_fd7287eb7f1365d9156e9da3ccb25196.pdf

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  • $\begingroup$ Do you mean $T^2=T\circ T$ ? $\endgroup$
    – Konstantinos Kanakoglou
    Commented Jan 25, 2019 at 21:10
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    $\begingroup$ Dually on algebras this would be an operator satisfying $T(x)T(y)=T(T(xy))$; an example is multiplication by a central element. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jan 25, 2019 at 21:43
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    $\begingroup$ Using the dual language written by @მამუკაჯიბლაძე (I'm sure this could be translated back easily, but prefer algebra to coalgebra): The equation implies that $T(C)$ is a (not necessarily unital) subalgebra. Letting $x = 1$, we have that $T(1) T(y) = T(T(y))$ (and similarly, $T(x) T(1) = T(T(x))$), so $T(1)$ is central within $T(C)$ and $T$ acts on $T(C)$ by multiplication by $T(1)$. $\endgroup$
    – user44191
    Commented Jan 28, 2019 at 4:31
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    $\begingroup$ Ali, may i ask for some details on the motivation of this question? In particular, i would like to know whether you are aware of some connection to physics? $\endgroup$
    – Konstantinos Kanakoglou
    Commented Feb 1, 2019 at 23:14
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    $\begingroup$ @KonstantinosKanakoglou To be honnest I did not have any physical motivation. Regarding mathematical motivation, the standard differentiation satisfies the functional equation for coalgebras. Now if we consider the "Algebra" analogy of this equation in the context of C* algebras we have automatic continiuity provided we have symmetric property. I wrote the motivations in a paper which link is available in my MO profile"on a functional equation for symmetric operator on C* algebras" $\endgroup$
    – Ali Taghavi
    Commented Feb 2, 2019 at 9:46

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About your first question:
Since you are asking for an example, take any group hopf algebra $k\mathbb{G}$, pick some subset $S\subset \mathbb{G}$ and denote $kS$ the linear subspace of $k\mathbb{G}$ generated by the set $S$. Let $T:k\mathbb{G}\rightarrow kS\subset k\mathbb{G}$ be the projection operator onto that subspace. Then your property is satisfied, by direct computation of both sides of your equation: $$ \Delta\circ T^2\big(\sum_{g\in\mathbb{G}}k_g g\big)=\sum_{g\in S}k_g g\otimes g=(T\otimes T)\circ\Delta\big(\sum_{g\in\mathbb{G}}k_g g\big) $$ since $T(g)=g$, if $g\in S\subset \mathbb{G}$ and $T(g)=0$, if $g\in G\setminus S$.

From a more general point of view, since $(f\otimes f)\circ \Delta=\Delta\circ f$ is -by definition- satisfied for any morphism of coalgebras, then your functional equation should be satisfied for any idempotent ( i.e. $T^2=T$) coalgebra endomorphism $T:C\rightarrow C$.

Edit: Attempting a translation of the comments to the OP, on the dual relation on algebras, by users : @მამუკა ჯიბლაძე and @user44191:
The functional equation stated at the OP implies that:
a). $\Delta\big(T^2(C)\big)\subseteq T(C)\otimes T(C))$ and
b). $T^2$ acts on $C$ as: $$T^2(c)=\varepsilon\big(T^2(c)_1\big)T^2(c)_2=T^2(c)_1\varepsilon\big(T^2(c)_2\big)= \\ =\varepsilon\big(T(c_1)\big)T(c_2)=T(c_1)\varepsilon\big(T(c_2)\big)$$ (you can get that by applying $Id\otimes\varepsilon$ and $\varepsilon\otimes Id$ on both sides of the functional equation at the OP $(T\otimes T)\circ \Delta=\Delta \circ T^2$).

However, i do not know about your second question in general.

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  • $\begingroup$ Thank you very much for your answer. $\endgroup$
    – Ali Taghavi
    Commented Jan 27, 2019 at 5:55
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    $\begingroup$ I think your translation is incorrect; removing the $T$ (analogous to where I essentially said $T(xy) \in C$), the inclusion goes in the other direction, that is, $(T \otimes T) \circ \Delta(C) = \Delta(T(T(C)) \subseteq \Delta(T(C))$. $\endgroup$
    – user44191
    Commented Jan 29, 2019 at 21:02
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    $\begingroup$ I've made a chat:chat.stackexchange.com/rooms/89079/… $\endgroup$
    – user44191
    Commented Jan 31, 2019 at 15:34
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    $\begingroup$ As a note, the example you have is in fact dual to a case of what @მამუკაჯიბლაძე wrote in the comments to the main post; the dual algebra to the group coalgebra is commutative, and the action you've chosen comes from multiplication by the idempotent $\mathbf{1}_S$ (which is central because the algebra is commutative). $\endgroup$
    – user44191
    Commented Feb 1, 2019 at 3:00
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    $\begingroup$ @ user44191: Nice! and maybe it would be even better, if all these notes and comments, including the OP and the answer, were recorded in the form of a new question here (towards particular or general solutions of the OP functional equation) or maybe in a post in nLab. $\endgroup$
    – Konstantinos Kanakoglou
    Commented Feb 2, 2019 at 2:02

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