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This is probably well-known to representation theorists, but this doesn't imply being well-known to me.

Let $k$ be a field, and let $A$ be a $k$-algebra that is finite-dimensional as a $k$-vector space. Let $S_1, S_2, \ldots, S_m$ be a complete list of pairwise non-isomorphic simple $A$-modules (up to isomorphism). Let $P_1, P_2, \ldots, P_m$ be the projective covers of $S_1, S_2, \ldots, S_m$, respectively. (Thus, $P_1, P_2, \ldots, P_m$ is a complete list of pairwise non-isomorphic indecomposable projective $A$-modules.) The Cartan matrix of $A$ is defined to be the $m\times m$-matrix in $\mathbb{Z}^{m\times m}$ whose $\left(i,j\right)$-th entry (for all $i$ and $j$) is the number of composition factors of $P_j$ isomorphic to $S_i$. (In other words, its $\left(i,j\right)$-th entry is $\dfrac{\dim \operatorname{Hom}_A \left(P_i, P_j\right)}{\dim \operatorname{End}_A \left(S_i\right)}$.) See §7.4 of Peter Webb's A Course in Finite Group Representation Theory for proofs and some background.

1. Is it true that the Cartan matrix of $A$ is invertible as a matrix in $\mathbb{Q}^{m\times m}$ ? (I think the answer is "no", but I don't know of a counterexample.)

2. Does this change if we require $k$ to be algebraically closed (or at least to be a splitting field for $A$) ?

3. Does this change if we furthermore require $A$ to be a Hopf algebra?

What I know is that the Cartan matrix of $A$ is invertible if $A$ is the group algebra of a finite group. This is proven in Corollary 10.2.4 of Peter Webb's above-mentioned book, but the proof uses the theory of Brauer characters, which as far as I know is particular to group algebras. (Or not? Is there a notion of Brauer characters for Hopf algebras? I certainly wouldn't find it strange, at least compared with the strangeness of classical Brauer character theory, but I have never seen such a notion.)

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    $\begingroup$ There is a four dimensional A providing a counterexample to both 1 and 2. Take a quiver with two vertices and two arrows, one in each direction between the two vertices. Let A be the quotient of the path algebra by the ideal generated by all length two paths. Then the Cartan matrix has all 1s. $\endgroup$
    – Peter McNamara
    Commented Feb 3, 2017 at 5:06
  • $\begingroup$ @PeterMcNamara: Ouch, should have thought of that one! Wondering whether there is a Hopf structure on it, though... $\endgroup$
    – darij grinberg
    Commented Feb 3, 2017 at 6:58
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    $\begingroup$ @darij: Brauer theory is rather special for the group algebra of a finite group, since it relies on having representations in characteristic 0 in the background; these are compared with representations in prime characteristic, which differ when the prime divides the order of the given finite group. So it's probably not realistic to look for close analogues when given an arbitrary finite dimensional Hopf algebra in prime characteristic. $\endgroup$
    – Jim Humphreys
    Commented Feb 7, 2017 at 18:57

2 Answers 2

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A counterexample is given by the Taft Hopf algebra. (I checked this at least for n=2, where this Hopf algebra also goes by a different name that I forget today).

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  • $\begingroup$ Ouch! This example was looking at me from the pages of the paper I'm writing with Vic and Jia that gave rise to this question, and yet I failed to notice it... $\endgroup$
    – darij grinberg
    Commented Feb 3, 2017 at 8:58
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    $\begingroup$ It actually gets even more extreme: a Hopf algebra over $\mathbb{C}$ is semisimple if and only if this matrix is invertible. In other words: for every finite dimensional non-semisimple Hopf algebra, the Cartan matrix is not invertible over $\mathbb{Q}$. $\endgroup$
    – Ehud Meir
    Commented Feb 3, 2017 at 9:25
  • $\begingroup$ The other name for $n=2$ is the "Sweedler Hopf algebra", the smallest noncommutative noncocommutative one. It also has some physical applications, as the Clifford algebra with degenerate metric (0,1). The nontrivial spinors on 2d Newton-Cartan geometry are exactly the projective covers of simple modules of this algebra. $\endgroup$
    – Lacia
    Commented Aug 3 at 10:54
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Perhaps it's worthwhile to mention one very large and natural class of finite dimensional Hopf algebras (over fields of prime characteristic $p$) for which the Cartan matrix is certainly not invertible: the restricted universal enveloping algebras of the Lie algebras of semisimple algebraic groups. Recall that if $G$ and hence its Lie algebra $\mathfrak{g}$ has dimension $d$, the restricted enveloping algebra has dimension $p^d$; it has for each ordered basis of $\mathfrak{g}$ an ordered basis of PBW type obtained by truncating at $p$th powers according to the $[p]$-operation on $\mathfrak{g}$. See for example the remarks at the end of section 4 in my 1971 paper here.

Take the simplest case, with $\mathfrak{g} = \mathfrak{sl}_2$. Here the $p$ simple modules have dimensions $1,2, \dots, p$. One block (indecomposable 2-sided ideal) involves just the simple/projective Steinberg module of dimension $p$, with Cartan invariant 1. Each other block involves just two simple modules of "linked" highest weights. If weights are identified with non-negative integers $0,1, \dots, p-1$ (with $p-1$ the highest weight of the Steinberg module), such blocks correspond to pairs of weights adding up to $p-2$. Then the Cartan invariants of the block are all 2.

Larger ranks lead to much more complicated (and not easily computed) Cartan invariants, but the theory developed in older work such as my paper yields the same result on the vanishing determinant of the matrix of these invariants.

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  • $\begingroup$ I don't even know what a Cartan invariant of a block is, but this looks really helpful, as we're always looking for examples of Hopf algebras in positive characteristic whose representations behave in interesting ways. Thank you! $\endgroup$
    – darij grinberg
    Commented Feb 3, 2017 at 21:33
  • $\begingroup$ @darij: The term "Cartan invariant" just refers to the multiplicity of a simple module in an indecomposable projective (PIM), so these are the entries in the Cartan matrix of the algebra. But they can be arranged in diagonal blocks, one for each block of the algebra. Thus the determinant of the Cartan matrix is the product of determinants of the various blocks. I was emphasizing that the restricted enveloping algebras here tend to have many blocks, in contrast to the corresponding finite groups of Lie type. $\endgroup$
    – Jim Humphreys
    Commented Feb 3, 2017 at 21:59
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    $\begingroup$ @darij: A comment on terminology. Apparently Elie Cartan did some work with finite dimensional associative algebras very early in his career, before going deeply into Lie groups, Lie algebras, and differential geometry. One result is two unrelated notions of "Cartan matrix": one for the matrix of Cartan invariants of an associative algebra (the 1962 Curtis-Reiner book is a good source, though lacking some of the extra topics and emphasis on homological algebra in their later 2-volume set), the other for the basic constants related to a semisimple Lie algebra. $\endgroup$
    – Jim Humphreys
    Commented Feb 3, 2017 at 23:27

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