Let $V$ be a finite-dimensional vector space and $$ U \subseteq W \subseteq V \otimes V $$ be a proper inclusion of vector subspaces. Then take the tensor algebra $$ T(V) = \bigoplus_{i=1}^{\infty} V^{\otimes i} $$ and the quotients $$ A_W = T(V)/\langle W \rangle, ~~~~~~~~~~~ A_U = T(V)/\langle U \rangle $$ by the ideals generated by $W$ and $V$ respectively. The quotients are again $\mathbb{Z}$-graded algebras (in fact quadratic algebras) and we have a surjective map from every homogeneous subspace of $A_U$ to every homogeneous subspace of $A_V$. At degree $2$ this is clearly not injective. However at higher degrees I would conjecture that it can be an isomorphism although I cannot see an example. I am most interested in the case where $A_U$ is a finite-dimensional algebra. In this can it happen that the non-zero homogeneous subspace of $A_U$ of highest order can have the same degree as the non-zero homogeneous subspace of $A_W$ of highest order?

  • $\begingroup$ Maybe I'm being dense, but as written you're just looking at free algebras generated by $V/W$ and $V/U$; also those presentations are not strictly speaking quadratic, because you have (albeit trivial in a sense, but still) relations in degree 1, not 2. $\endgroup$
    – Denis T
    Commented May 25, 2023 at 22:12
  • $\begingroup$ @Denis: You are not being dense - this is a typo. I have fixed it now. $U$ and $W$ should be in $V \otimes V$. Thanks for pointing this out! $\endgroup$ Commented May 25, 2023 at 22:15

1 Answer 1


There are plentiful examples of $m$-generated quadratic algebras $A$ with $A_3 = 0$, and dimension of relation subspace $R$ being as low as $m^2/2$. One of those algebras is $$\Bbb k\langle x_1, y_1, \dots, x_n, y_n \rangle / (x_i y_j, x_i x_j - y_i y_j).$$ Moreover, if $\operatorname{dim} R \geq 3m^2/4$, then for very generic (i. e. lying in intersection of countably many certain Zariski opens in Grassmanian) choice of $R$ corresponding quadratic algebra will have $A_3 = 0$ and be Koszul.

Obviously, any proper quadratic quotient of those will give you an example of epimorphism of quadratic algebras such that their Hilbert series only differ in degree 2. You can take direct sum with your favourite quadratic algebra to have something in degrees $\geq 3$ if you want.

UPD: I need to do a due diligence and add reference. I can wholeheartedly recommend to anyone interested in quadratic algebras and combinatorial algebra in general to read small book "Quadratic algebras" by A. Polishchuk and L. Positselski. https://bookstore.ams.org/ulect-37

  • $\begingroup$ Thanks a lot for the answer. For the example you gave I am having trouble seeing that $A_3 = 0$. Why should something like $x_1x_2x_3$ be equal to zero? Also, what do you mean by $m^2/2$? I am sure this is standard notation, but I am not familar with it. $\endgroup$ Commented May 26, 2023 at 12:45
  • $\begingroup$ If a trinomial has no $x_i y_j$ substring, then it is one of four types: $XXX, YXX, YYX, YYY$. Using relations $XX = YY$, you can reduce them all: $YYX \mapsto XXX \mapsto XYY = 0$, $YXX \mapsto YYY \mapsto XXY = 0$. $m$ is the number of generators, that's written in the first line of the answer. $\endgroup$
    – Denis T
    Commented May 26, 2023 at 15:01
  • $\begingroup$ I see, now it is clear that $A_3$ vanished. Thanks! For the notational question: I was asking what $m^2/2$ denotes. Is it the number of elements in $\{g^2 ; g \in m\}$ divided by $2$? $\endgroup$ Commented May 26, 2023 at 17:18
  • $\begingroup$ It denotes a number $m^2/2$. $\endgroup$
    – Denis T
    Commented May 26, 2023 at 22:48

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