(Original title: is time-odering operator a linear operator?)
I'm confused with two formulas, one of which is $$ \mathcal{T} \exp \left [-\frac{\mathrm{i}}{\hbar} \int_{t_0}^t \mathrm{d} t' \hat{H}_I(t') \right] =\sum_{N=0}^\infty\frac{1}{N!} \left (-\frac{\mathrm{i}}{\hbar} \right)^N \int_{t_0}^t \mathrm{d} t_N \int_{t_0}^t \mathrm{d} t_{N-1} \cdots \int_{t_0}^t \mathrm{d} t_1 \mathcal{T} \hat{H}_I (t_N) \hat{H}_I(t_{N-1}) \cdots \hat{H}(t_1). $$ I'm clear about how the right hand side is derived following this page, but I don't understand how it goes to the left. Is that just a definition? Otherwise I have to expand the exponential to series and admit the time ordering operator is linear so that $\mathcal T(A+B)=\mathcal T(A)+\mathcal T(B)$. But I haven't seen any description on that.
The other one is $$ \mathcal T([A,B])=\mathcal T(AB-BA)=\mathcal T(AB)-\mathcal T(BA)=0 $$ whether $[A,B]$ is $0$ or not. To make the second equal sign hold, $\mathcal T$ is again required to be linear. But if it is linear, and $[A,B]$ is a constant, say $c\not =0$, $$ \mathcal T([A,B])=\mathcal T(c)=T(cI^2)=c\mathcal T(I^2)=c\not=0, $$ which is a contradiction.
This answer points out that $\mathcal T$ maps on an operator-valued function of time in a space with more than one dimension, but no information on the linear property is mentioned.
Later edited:
$[A,B]$ doesn't have to be a constant, in fact it's impossible because $A,B$ are operators dependent on different time variables. But the paradox survives if $[A,B] = c1\!\!1$, where $c=c(t,t')$ is a function of the time variables and $1\!\!1$ denotes the identity.
For example, commutator of the position and momentum in Heisenberg picture writes, $$ [x(t),p(t')]=\cos(t-t')=\cos t\cos t'+\sin t\sin t'=C(t)C(t')+S(t)S(t') $$ except the coefficients, where $$ C(t)=\cos t1\!\!1,\quad S(t)=\sin t1\!\!1. $$ Therefore (suppose $t>t'$) $$ 0=\mathcal T([x(t),p(t')]) =\mathcal T\left[C(t)C(t')+S(t)S(t')\right]= C(t)C(t')+S(t)S(t')=\cos(t-t'). $$ Contradiction again.
I thought the point is, time-ordering operation on product of operators--namely, monomial--is well-defined, but operation on sum of these products, which is polynomial of operators, is not.
I thougth the polynomial can be devided into monomials that yields the same result, but according to the above example, that's not true.
Thanks for answering! I'm not content with just regarding it as a notation, since I can't tell where paradoxes like above lie.