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What is the right definition of the symmetric algebra over a graded vector space V over a field k?

More generally: What is the right definition of the symmetric algebra over an object in a symmetric monoidal category (which is suitably (co-)complete)?

Two possible definitions come to my mind:

1) Take the tensor algebra over V and identify those tensors which differ only by an element of the symmetric group, i.e. take the coinvariants wrt. the symmetric group. The resulting algebra A is then the universal algebra together with a map V -> A such that the product of elements in V is commutative.

2) Take the tensor algebra over V and divide out the ideal generated by antisymmetric two-tensors. In this case, the resulting algebra A is the universal algebra together with a map V -> A such that the product of A vanishes on all antisymmetric two-tensors (one could say that all commutators of A vanish).

The definition 1) looks more natural and gives, for example, the polynomial ring in case V is of degree 0.

The definition 2) applied a vector space shifted by degree 1 gives (up to degree shift) the exterior algebra over the unshifted vector space. However, in characteristic 2 for example, one doesn't get the polynomial ring if one starts with a vector space of degree 0.

Finally, both definitions have a shortcoming in that they don't commute well with base change.

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Symmetric algebras (aka free commutative associative unital algebras) are given by a functor, and they satisfy a universal property: If M is a module over a commutative ring k and R is a commutative k-algebra, then k-algebra homomorphisms from Symk(M) to R are in bijection with k-module maps from M to R. This bijection should be functorial with respect to R (i.e., ring homomorphisms in the target). More succinctly, the symmetric algebra functor is left adjoint to the forgetful functor from commutative k-algebras to k-modules. This is the description in the "categorical properties" section of the Wikipedia article.

The universal construction normally yields definition 1, but in the derived world, you might have to do something extra, like take homotopy coinvariants (this means taking some kind of resolution to get a free symmetric group action).

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My understanding is that the "right" way to define the symmetric algebra comes from a braiding that tells you how the symmetric group acts on tensor products. And an easy and general way to get such a braiding is to consider the category of representations of a quasi-triangular Hopf algebra; this is a short way to define the category of supervector spaces and recover the usual notion of graded-commutativity there.

The problem is that when you say "graded" (say, with respect to a group) you're only specifying at best a Hopf algebra, not a quasi-triangular structure. So the answer depends on what possible quasi-triangular structures are floating around (in the $\mathbb{Z}/2\mathbb{Z}$ case on the group algebra of $\mathbb{Z}/2\mathbb{Z}$) and of course for a group of order $|G|$ terrible things are going to happen in characteristic dividing $|G|$.

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  • $\begingroup$ So you are saying that the usual definition of a graded vector space only makes sense away from 2, right? $\endgroup$
    – Marc Nieper-Wißkirchen
    Commented Nov 29, 2009 at 10:40
  • $\begingroup$ Right. My guess is that one should be able to substitute a different Hopf algebra in characteristic 2 that allows you to recover squaring odd elements to zero, but I haven't really thought about it. $\endgroup$
    – Qiaochu Yuan
    Commented Nov 29, 2009 at 14:23
  • $\begingroup$ In any characteristic, the right Hopf algebra should be the functions on the group scheme of second roots of unity as this also makes sense over the integers. This still leaves the question on how to define the braiding over the integers. $\endgroup$
    – Marc Nieper-Wißkirchen
    Commented Nov 30, 2009 at 8:36
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In simple terms, it seems to me to depend on your definition of "graded commutative k-algebra". I would take this mean that

$xy =(-1)^{|x||y|}yx$

where |x| denotes the degree of the homogenous element x. So I would divide by the ideal generated by

$xy -(-1)^{|x||y|}yx$

for homogeneous elements x, y in V. This does capture the polynomial algebra if the vecotr space is in degree 0, and the exterior algebra if the vector space is in degree 1 and the characteristic is not 2, but the polynomial algebra if the characteristic is 2. This is the right definition for algebraic topology.

But of course, there are other possible definitions, as others have said.

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  • $\begingroup$ Could you clarify on in which sense your definition is the right definition for algebraic topology, e.g. by giving an example where an exterior algebra shows up when considering, say, Z coefficients but a polynomial algebra, when Z/(2)-coefficients are used. Thanks! $\endgroup$
    – Marc Nieper-Wißkirchen
    Commented Nov 29, 2009 at 10:23
  • $\begingroup$ @Marc: Well, take the classifying space of Z/p. For p=2, this is infinite dimensional real projective space, whose cohomology with Z/2 coefficients is a polynomial ring on a generator of degree 1. For p odd, it is an infinite dimensional lens space whose cohomology ring with Z/p coefficients is an exterior algebra on a generator of degree 1 tensor a polynomial algebra on a generator of degree 2. The difference between these two cases is precisely that you cannot have an polynomial algebra on a generator of degree 1 if p is odd. Don't know if this completely answers you. $\endgroup$
    – Mark Hovey
    Commented Dec 1, 2009 at 15:11
  • $\begingroup$ Thanks, Mark. This answers my question on your original answer. $\endgroup$
    – Marc Nieper-Wißkirchen
    Commented Dec 15, 2009 at 11:55
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The right definition is: take the free associative (tensor) algebra generated by $V$; divide out the ideal generated by the elements $xy-(-1)^{|x||y|}yx$ for all homogeneous $x$, $y\in V$ and $z^2=0$ for all odd $z\in V$. This should commute with the base change well (when $V$ is flat over your base).

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  • $\begingroup$ For graded vector spaces V = V_even + V_odd this yields the classical symmetric algebra over V_even tensored with the classical exterior algebra over V_odd, doesn't it? However, does this definition make sense in the context of general symmetric monoidal categories? $\endgroup$
    – Marc Nieper-Wißkirchen
    Commented Nov 30, 2009 at 8:34
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    $\begingroup$ Yes it does. No, it apparently doesn't: for in characteristic 2, the permutation operators do not depend on the parity, so they cannot distinguish between the symmetric and exterior generators. This may be a common situation for superstructures in small characteristic: e.g., in the definition of a Lie superalgebra one has to pay attention not only to char=2, but also to char=3. The correct definition requires, in addition to the Lie bracket, the data of a "squaring" map q: g_odd -> g_even. When 2 is invertible in your ground ring, this problem does not present itself. $\endgroup$
    – Leonid Positselski
    Commented Nov 30, 2009 at 11:06
  • $\begingroup$ For a $G$ graded vector space, where $G$ is a finite abelian group, what should be the ideal that we consider ? Of course one considers the elements of the form $xy-(-1)^{|x||y|}$. Here there is no $\mathbb Z_2$-grading and hence no odd elements, what kind of elements would replace $z^2$ ? I would like that it should commute with the base change. $\endgroup$
    – jack
    Commented Sep 26, 2023 at 10:03
  • $\begingroup$ What does $(-1)^{|x||y|}$ mean when $|x|$, $|y|\in G$ and $G$ is a finite abelian group? $\endgroup$
    – Leonid Positselski
    Commented Sep 26, 2023 at 20:27
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This is not an answer, as I think Scott did a better job than I could have. Another algebra that generalizes the symmetric algebra in characteristic non-zero is the algebra of divided polynomials. Let V be a finite-dimensional vector space over k a field, and let V* be its dual space. For each n, write the space of n-linear maps from V* to k, and take the subspace of maps that are invariant under the natural Sn-action. The direct sum of all of these spaces is a k-algebra, where the multiplication is as follows. To multiply a symmetric m-linear map by a symmetric n-linear map, for each subset of size m of a set of size m+n consider the (m+n)-linear map that sends the subset through the m-linear multiplicand and the rest through the n-linear one; then add up all the m-choose-n many ways to do this. Then this algebra agrees with the symmetric algebra over V when k is characteristic-zero, but not otherwise. For example, when V is one-dimensional with basis vector x, then the symmetric algebra is the polynomial algebra k[x], whereas the above construction yields the algebra k[x,x2/2,x3/6,x4/24,...]. In fact, the symmetric algebra of V* and the above algebra are both naturally graded Hopf algebras, and in fact they are duals as graded Hopf algebras.

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  • $\begingroup$ I think the PD algebra can also be viewed as the ring of translation-equivariant algebraic differential operators on V. Then the duality of algebras you mentioned has something to do with the way both of them fit into the Weyl algebra. $\endgroup$
    – S. Carnahan ♦
    Commented Nov 29, 2009 at 4:37
  • $\begingroup$ Possibly. If V is an algebraic group over k, are the left-translation equivariant differential operators then the divided-power universal enveloping algebra of Lie(<i>V</i>)? $\endgroup$
    – Theo Johnson-Freyd
    Commented Nov 29, 2009 at 6:59
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Which one is the right definition depends on what you want to do with it.

No matter how much technology you throw at the question, including homotopy coinvariants and quasi-triangular Hopf algebras, "right" is a relative notion :D

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