$\newcommand{\dual}{\mathrm{dual}}\DeclareMathOperator\End{End}\DeclareMathOperator\Fun{Fun}\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\GL{GL}$Around 1986–7 Yu.Manin proposed natural and simple construction how to associate with algebra $A$ another algebra $\End(A)$ which is kind of non-commutative symmetries of $A$, or kind of $\Fun(\GL(\Spec(A))$, with prime applications to construct quantum groups in that way. Decades later the construction gained applications in more classical representation theory and knot theory ("Manin matrices)" (see also MO-count-Fp, MO-Koszul-1).
Let me first ask the question and remind the construction afterwards – it is so natural and simple that can be described in few words.
Question 1: Assume $A$ is quadratic Koszul algebra, is it true that $\End(A)$ is also Koszul algebra ?
Thanks to Steven Sam we know a positive answer to the question for the most basic example $A=k[x_1,x_2]$ and so $\End(A)$ are standard Manin matrices. It is very natural to expect that is would be true for $A=k[x_i]$ and q-super-analogs $A=k[x_i,\psi_j]$ – which proved to satisfy so many remarkable properties, that cannot believe that Koszulity is not one of them. The case of general quadratic algebras is less studied, but optimistically we might hope even for that.
Quadratic algebras are naturally graded, and so it is interesting to study their Hilbert series.
Question 2: Assume Hilbert series for $A$ and its quadratic dual are known , is there any way to express Hilbert series for $\End(A)$ in their terms or produce Hilbert series of $\End(A)$ in some other way from $A$ and $A^{\dual}$ ?
Again in the basic example $A=k[x_1,x_2]$ Steven Sam provided a Hilbert series. Naively we may think that $\End(A) = A\otimes A^{\dual}$ so very naively the Hilbert series should be related to the product of these two, it seems it is not the case – we are dealing with non-commutative symmetries and that is not so simple, but it seems natural to search for some kind pattern which will govern Hilbert series of $\End(A)$ since it is not a general algebra but quite specific.
Reminder on Manin's construction.
Motivation: consider group $G$ acting on some space $H$, that is we have a map: $G\times H \to H$ . Consider algebras of functions – se we have a map: $\Fun(H) \to \Fun(G) \otimes \Fun(H)$, which satisfies natural analogues of co-associativity. But that language of algebras works for general algebras, so we proceed as follows:
Conceptual coordinate free definition: Hopf algebra $\End(A)$ is defined as universal co-acting algebra on $A$, i.e. there is map $A \to \End(A) \otimes A$ which satisfies natural co-action axioms ( analogs of co-associativity). Where "universal" means that we do not impose any other relations except those which follow the co-action requirement i.e. it is in a sense "largest" co-acting algebra. (That part might sound not so clear if ones hears for the first time, but the "down-to-earth" wording below will clarify that.)
Simple words, concrete approach:
Choose some generators $x_i$, $i=1,…,n$ for algebra $A$, then $\End(A)$ is defined by a set of $n^2$ generators $M_{ij}$ for $i,j=1,…,n$, and the relations appeared like that:
consider $y_{i} = \sum_j M_{ij}x_j $ , and require $y_i$ satisfy exactly the same relations as $x_i$. So in the matrix language – apply matrix $M$ to vector $x$ and require result $Mx$ satisfy the same relations as vector $x$.
For example for $A = k[x_1,x_2]$:
$$\begin{pmatrix} y_1 \\\ y_2 \end{pmatrix} = \begin{pmatrix} a & b \\\ c & d \end{pmatrix} \begin{pmatrix} x_1 \\\ x_2 \end{pmatrix} $$
So the only requirment $[y_1,y_2]=0$, or explicitly $ 0 = [ax_1+bx_2,cx_1+dx_2] =[a,c]x_1^2+([a,d]-[c,b])x_1x_2 + [b,d]x_2^2$, thus we get the three relations $[a,c]=0, [b,d]=0, [a,d] = [c,b]$.
Comultiplication is standard $\Delta(M_{ij}) = \sum_k M_{ik}\otimes M_{kj}$ – that is just dual to the standard matrix multiplication formula.
Surprises. There are two surprises here – first we started with plain commutative algebra of polynomials – but what we get is non-commutative algebra of "symmetries". Second if work with matrices $M$ then despite huge non-commutativity of their elements – "all" the linear algebras works for such matrices – one can define determinant, Cramer inversion formula works, Cayley-Hamilton, Newton … all the identities are working despite non-commutativity and in some sense that is precisely the largest possible non-commutativity which allows linear algebra to work, in general matrices with non-commutative entries behave badly.
Generalizations
In the same way one may consider "non-commutative morphisms" between the two algebras $A$, $B$ – to get "rectangular" matrices. And there are braided generalizations – when we compute relations between $Mx$, $Mx$ we may not assume that $M$ and $x$ commute, but require some braided generalization.
We may ask the same questions for these cases.
PS
There is nice MO-question by Mare "Question 1: Is there a computer algebra system that can check whether such an algebra is Koszul?" and comment by
Leonid Positselski "I think that Question 1 is long known to be algorithmically unsolvable. ". So it might be Koszul duality questions are not that much easy.
Best Answer
Q1: This algebra is just the Manin black product of $A$ and $A^!$ (in other words, the Koszul dual of the Segre product of $A$ and $A^!$), and hence it is Koszul. (As requested, the Segre product of two graded algebras $A$ and $B$, is the algebra whose graded components are $A_n\otimes B_n$.) The proof of the fact that the Segre product of two Koszul algebras is Koszul is, for instance, in Chapter 3 of the book of Polishchuk and Positselski.
Q2: In particular, the Hilbert series of the Segre product is clearly the Hadamard product of the series of $A$ and $A^!$, so the series of your algebra is, up to alternating signs, the multiplicative inverse of the latter. For example, in the case considered by Steven Sam in the MO question you link, the Hilbert series of $A$ (polynomial algebra) is $1+2t+3t^2+O(t^3)$, the Hilbert series of $A^!$ (exterior algebra) is $1+2t+t^2$, so the Hadamard product of these series is $1+4t+3t^2$, and so the Hilbert series of your algebra is $\frac{1}{1-4t+3t^2}$, as indicated by Steven.