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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 t …

The answer for $L_1=\sum_ix_i^2\partial_i$ can be derived in a rather straightforward way (I changed your convention a little bit to match the usual formulas for Virasoro algebra). Namely, use the det …

I accidentally (looking for something else) came across another paper where a very elegant explanation is given: Dan Laksov, Anders Thorup: Schubert Calculus on Grassmannians and Exterior Powers India …

This is very standard. For a compact complex variety admitting a cell decomposition, the (co)homology is the free Abelian group generated by the cells (over $\mathbb{C}$ there is no room for the diffe …

In your context, you want to think of the Schur-Weyl duality as a way to construct representations of $GL(V)$ out of representations of symmetric groups. To give a precise answer along these lines tha …

There is a $q$-binomial identity that I encountered in one paper I am reading (https://arxiv.org/abs/1910.06193) which probably admits a very simple proof that I do not see: for two nonnegative intege …

I recall seeing in various sources the terminology "cover preserving embedding" and "cover preserving subposet". Googling it now (https://www.google.com/search?q=poset+%22cover+preserving%22) brings s …

The argument goes as follows. Let us consider the events $A_i=\{ i(i+1) \text{ occurs in a permutation} \}$ and $B_i=\{ (i+1)i \text{ occurs in a permutation} \}$. Some pairs of events like that canno …

Now, to the matter of "positive formulas". Classical result of Kraskiewicz and Weyman (preprint W. Kraskiewicz, J. Weyman. Algebra of Invariants and the Action of a Coxeter Element. Math. Inst. Copern …

For your question about Lie triple systems: In my article "Veronese powers of operads and pure homotopy algebras" joint with Martin Markl and Elisabeth Remm, Eur. J. Math. 6 (2020), 829-863 (https://l …

For rather trivial reasons, $$ 1+\sum_{(k,l)\ne (0,0)}p(k,l)x^ky^l=\prod_{(p,q)\ne(0,0)}\frac{1}{1-x^py^q} . $$ Since these numbers include, as $p(n,0)$, the one-dimensional partition numbers, you …

No, it is not multiplicity-free. Already for $d=6$, this representation contains the Schur functor $S^{4,4,4,2,2,2}$ twice. This can be easily checked in Magma (even the online calculator) issuing the …

The answer (already given in comments, with a small misprint/mistake) is: (a) for even $r=2k+2$, it is $q^{m-1}+(q-1)q^{m-k-2}\eta((-1)^{k+1})$, where $\eta$ is the quadratic character of $\mathbb{F} …

To offer a slightly more geometric viewpoint on the same, the space $\bigoplus_q \mathop{\mathrm{Hom}}_K(\Lambda^q(\mathfrak{p}),\mathbb{C})$, which is the direct sum of all spaces you are considering …

This is inspired by Fedor's answer: Consider $$ f(m):=-\frac{\binom{ab}{m}\binom{cd}{m}(m-bc)(m-ad)}{\binom{ad-1}{m}\binom{bc-1}{m}(a-c)(b-d)}. $$ Then $$ f(m+1)-f(m)=-\frac{\binom{ab}{m+1}\binom …