# 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). Define for $$k,t \in \{1,...,n \}: m_{k,t}:= t-max(0,t-k)=min(k,t)$$ (so $$m_{k,t}=t$$ for $$k>t$$ and $$m_{k,t}=k$$ for $$t \geq k$$).

Write $$S= \{s_1,...,s_{r-1} \}$$ with the $$s_i$$ in increasing order and $$r-1$$ the cardinality $$|S|$$ of $$S$$ and set $$s_r:=n$$.

Let $$a_{i,j}:=m_{s_i,s_j}=min(s_i,s_j)$$ for $$i,j \in \{1,...,r\}$$. Let $$A_S$$ be the matrix with entries $$a_{i,j}$$, called the Cartan matrix of $$S$$ and let $$d_S$$ be defined as the determinant of $$A_S$$, called the Cartan matrix of $$S$$.

We set $$A_{\emptyset}=n$$ and thus $$d_{\emptyset}=n$$.

Question 1: Is there an explicit formula for $$d_S$$?

Question 2: I can prove that $$d_S$$ is non-zero. Is it more general true that $$d_S >0$$?

Now call $$S$$ great in case the following holds.

-In case $$n$$ is even, $$|S|$$ is odd,$$min(S)$$ is odd and $$max(S)$$ is odd and all differences $$s_i-s_{i-1}$$ are odd for $$i=2,...,r-1$$.

-In case $$n$$ is odd, $$|S|$$ is even, $$min(S)$$ is odd, $$max(S)$$ is even and all differences $$s_i-s_{i-1}$$ are odd for $$i=2,...,r-1$$.

-$$\emptyset$$ is great if and only if $$n$$ is odd.

Question 3: Is it true that $$d_S$$ is odd if and only if $$S$$ is great? Are the great subsets $$S \in \{1,...,n-1 \}$$ enumerated by the Fibonacci numbers $$F_n$$?

Question 2 and 3 have a positive answer for $$n \leq 14$$ and thus there is good evidence that it is true and something nice is going on. I can prove (looking at entries mod 2) that if $$S$$ is great, then $$d_S$$ is odd, but Im not sure about the converse. The result being so nice, I would think that there is a short proof that avoids heavy computations.

Now here come some more weird observations that might be much harder to show.

Question 4: Call a subset $$S$$ prime in case $$d_S$$ is prime. When is $$S$$ prime? For $$n \leq 20$$ it was true that the number of prime $$S$$ is given by $$\sum\limits_{k=1}^{m}{\pi(k)}$$, where $$\pi(k)$$ is the prime counting function. See https://oeis.org/A046992 . Is this true in general?

Some other findings that occured in the oeis:

The number of subsets $$S$$ with $$d_s=1 mod 3$$ seems to be given by the seqeunce https://oeis.org/A075111 , related to the tribonacci numbers. Is this true?

The number of subsets $$S$$ with $$d_S=1 mod 4$$: https://oeis.org/A005252 .

With $$d_S=3 mod 4$$: https://oeis.org/A024490.

Note that all the 3 previous sequences are related to the Fibonacci sequence, which might indicate that this all has a secret connection to Fibonacci combinatorics (which I can not see at the moment).

The number of $$S$$ with $$d_S$$ such that $$Gcd(d_S,n)=1$$ seem to be given by https://oeis.org/A100347.

Representation theoretic background.

Let $$A=A_n=K[x]/(x^n)$$. Then there is a unique indecomposable $$A$$-module $$M_i$$ of dimension $$i$$ for $$i=1,...,n$$.

The subsets $$S \subseteq \{1,...,n-1 \}$$ are in bijection with the (faithful) module $$M_S := A \oplus \bigoplus\limits_{i \in S}^{}{M_i}$$. Then $$d_S$$ is the Cartan determinant of the algebra $$End_A(M_S)$$.

• Am I right that $m_{k,t}=\min(k,t)$? – Ilya Bogdanov Nov 25 '19 at 13:45
• @IlyaBogdanov Yes, thanks :D I wrote down the formula directly as computated from Hom-spaces. But of course the simplification in brackets is equal to min(k,t). I added this to make it more clear. – Mare Nov 25 '19 at 13:46

If $$s_1, the matrix $$A_S$$ is the matrix of the quadratic form $$s_1(x_1+\ldots+x_r)^2+(s_2-s_1)(x_2+\ldots+x_r)^2+\\(s_3-s_2)(x_3+\dots+x_r)^2+\ldots+(s_r-s_{r-1}) x_r^2.$$ This quadratic form is obviously positive-definite which already implies $$d_S>0$$. Next, we get $$(A_Sx,x)=(D_STx,Tx)\Rightarrow A_S=T^tD_ST,$$ where $$D_S$$ is diagonal with elements $$s_1,s_2-s_1,\ldots,s_r-s_{r-1}$$ and $$T$$ is the triangular map $$x\to (x_1+\ldots+x_r,x_2+\ldots+x_r,\ldots,x_r)$$. Thus $$\det A_S=\det D_s=s_1(s_2-s_1)(s_3-s_2)\ldots (s_r-s_{r-1})$$ which answers other questions.
For any set $$S=\{s_1,\dots,s_k\}\subset[0,\infty)$$ with $$s_1<\dots, the matrix $$A_S=(\min(s_i,s_j))_{i,j=1}^k$$ is the covariance matrix of the random vector $$(B(s_1),\dots,B(s_k))$$, where $$B(\cdot)$$ is a standard Brownian motion. So, $$A_S$$ is positive semidefinite. Moreover, since the random variables $$B(s_1),\dots,B(s_k)$$ are clearly linearly independent, $$A_S$$ is positive definite, and hence $$\det A_S>0$$.