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4 votes
0 answers
163 views

An identity for Schur polynomials

Given a partition $\lambda$, the Schur polynomials can be defined, among many other ways, as $$S_{\lambda}(\xi_1,\dots,\xi_a)=\frac{\det\left(\xi_i^{\lambda_j+a-j}\right)_{i,j=1}^a}{\det\left(\xi_i^{a-...
T. Amdeberhan's user avatar
2 votes
2 answers
214 views

Cartan determinants of subsets

Let $n \geq 3$ be fixed. We associate to every subset $S \subseteq \{1,...,n-1 \}$ a number, which we call Cartan determinant of $S$ (see the end of this post for a representation theoretic background)...
Mare's user avatar
  • 26.5k
7 votes
1 answer
462 views

On a problem for determinants associated to Cartan matrices of certain algebras

This is a continuation of Classification of algebras of finite global dimension via determinants of certain 0-1-matrices but this time with a concrete conjecture and using the simplification suggested ...
Mare's user avatar
  • 26.5k
11 votes
2 answers
558 views

Classification of algebras of finite global dimension via determinants of certain 0-1-matrices

I restrict to the elementary problem that is equivalent to give a classification when Morita-Nakayama algebras have finite global dimension (see the end of this post for some background). A Morita-...
Mare's user avatar
  • 26.5k
13 votes
1 answer
399 views

Is there a Giambelli identity with dual representations?

For natural numbers $a,b$ with $b\leq n-1$, let $V_{ (a|b)}$ be the irreducible representation of $GL_n$ with highest weight vector $(a+1, 1^b, 0^{n-b-1})$ where the exponentiation denotes repetition. ...
Will Sawin's user avatar
  • 148k
1 vote
1 answer
96 views

Realising matrices as Cartan matrices

Given a matrix with natural numbers $\geq 0$ as entries and having determinant equal to one and positive diagonal entries. Is it the Cartan matrix of a finite dimensional algebra of finite global ...
Mare's user avatar
  • 26.5k
6 votes
0 answers
375 views

Monomial base change and the Vandermonde

Denote the falling factorials by $(x)_k=x(x-1)\cdots(x-k+1)$. The Vandermonde determinant is given by $\det\left[x_i^{j-1}\right]_1^n=\prod_{i<j}(x_j-x_i)$. It is well-known that in as much as ...
T. Amdeberhan's user avatar
6 votes
1 answer
208 views

Sum identities with immanants

For $\chi$ being an irreducible character of the symmetric group $S_n$ and being $M$ a complex $n\times n$-matrix, I would like to show $$ \sum_{\sigma, \rho \in S_n} \overline{\chi(\sigma)} \chi(...
Malte's user avatar
  • 93
8 votes
0 answers
488 views

det(A)det(B) = det(AB+correction), Capelli identities, "factorized" representation of $\mathfrak {gl}_n$

Context: Some probably know that there are Capelli identities which state $$det(A)det(B) = det(AB+correction)$$ for some matrices with non-commuting elements, they go back to the 19-th century, but ...
Alexander Chervov's user avatar