Wait what? Not closed under multiplication by $-1$? I always thought they were! Can you give an example of where they are not?

The second issue of the resulting algebra being unital or not is trickier (that is why my two comments are comments rather than answers) but I am also not sure if we agree on what that word means. I thought it means there must be an element $e \in V$ such that $ex = xe = x$ for all $x \in V$. But this doesn't hold in the case of Lie algebras, which seems to be one of your examples

Any algebra $V$ can be viewed as a map (the multiplication) $\mu: V \otimes V \to V$. If we don't require associativity OR any relation to $G$, then of course we can define an algebra structure on every irrep $V$ of $G$. Of course you do want some relation to $G$, I expect you want $\mu$ to be a $G$-map? Then the first step is to find out (using Weyl character formula or such) for which $V$ we have that $V$ appears as a summand in the decomposition of $V \otimes V$ into irreps. This is already a restriction.

Why the downvote?

O this is really nice!

Very interesting question by the way

Also in my first comment I meant $\mathfrak{q}_\mathbb{C}$ where I wrote $\mathfrak{q}_\mathfrak{C}$ but it is too late to edit now.

About maximally non-compact: the Cartan subalgebra from which $\mathfrak{b}$ is built in the Knapp-world is supposed to be the complexification of the Cartan subalgebra $\mathfrak{t} \oplus \mathfrak{a}$ of $\mathfrak{g}$, with $\mathfrak{t}$ a Cartan subalgebra of $\mathfrak{m}$ and with $\mathfrak{m}$ and $\mathfrak{a}$ as in your MSE question. Now intuitively it seems to me that the fact that $\mathfrak{t} \oplus \mathfrak{a}$ is maximally non-compact is the direct result of $\mathfrak{a}$ being maximal itself (subject to some conditions) as part of its definition. So it seems unavoidable.

Doesn't the minimality condition (combined with your definition of parabolic subgroup) imply that $\mathfrak{b}$ should equal $\mathfrak{q}_\mathfrak{C}$ (if and when such $\mathfrak{q}$ exists) rather than being merely contained in it? Am I missing something?

Singularities are called regular if they are of order less than or equal to the degree of the equation itself

Generally speaking we say that a differential equation $g'(u) = g(u)/f(u)$ for some fixed polynomial $f$ has a singularity at $u = 0$ of order equal to the deg $f$. Here this order is 2. And the singularity takes place at $u = 0$ or equivalently at $x$ equals infinity.

@Annix The notation $y' = y$ obscures with respect to what variable the derivative is taken, but let's say it is with respect to a variable $x$. Now to see what happens with solution to this equation 'at infinity' we look at what happens at $u = 0$ for a variable $u$ defined as $u = 1/x$. Suppose that $y$ is a function of $x$ that satisfies $y' = y$ and we define $g(u) = y(1/u)$ then what differential equation does $g$ satisfy? By the chain rule $g$ is a solution to $g'(u) = (1/u^2)g(u)$. (ctd in next comment)

The last comment clarifies the question a lot, perhaps add it to the 'main' body of the question?

O wait, I see you said something about that in the comment to the other answer: at least before 1915

What year is this?

@AFK It is an old post, but it was linked to in a reason MSE post, that's why I got here. Can you elaborate a bit on your cohomological reasoning in the post? Even if the below answer is correct and this injectivity does not prove the result, I would still be curious to understand better how it relates to zeta values in the first place

The addition of the third term is a great improvement over standard 'and' but for the reasons pointed out above the 'it would be possible' symbol is a bit too weak. In an ideal world I guess we would replace it by a symbol that says 'but conditioned on $X$, not $Y$ it is more probable than $Y$'. If only such a symbol existed...