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The rule $(v_1,w_1)⋅(v_2,w_2)=(v_1+v_2,w_1+w_2+(v_1∧v_2))$ defines a group structure on the vector space $V⊕(V∧V)$ whenever $V$ is itself a vector space over some field $F$.

What is a more common name for this group?

  • The group is nilpotent of class at most $2$, with commutator $[(v_1,w_1),(v_2,w_2)] = (0,2(v_1∧v_2))$.
  • The group has exponent $\operatorname{char}(F)$ since $(v_1,w_1)^n = (v_1^n,w_1^n)$.
  • When $\operatorname{char}(F) = 2$, this is just an elementary abelian $2$-group.
  • When $\dim(V) = 1$, this is just the abelian group $V$, so has a faithful $2$-dimensional F-module.
  • When $\dim(V) = 2$ and $\operatorname{char}(F)≠2$, this is a maximal unipotent subgroup of $\operatorname{GL}(3,F)$, so has a faithful $3$-dimensional $F$-module.

I don't recognize it when $\dim(V) = 3$.

When $\dim(V) = 3$ and $\operatorname{char}(F)≠2$, does the group have a faithful $F$-module of dimension independent of $F$?

Is this a maximal unipotent subgroup of some classical group or does it otherwise have a standard description as a matrix group?

I am primarily concerned with the case that $F$ is a finite prime field (and am hoping to do better than a permutation module of dimension $|F|^5$), but I assume many sorts of answers should be mostly independent of the field.

I don't have much hope for large $\dim(V)$, since the nilpotency class of the group remains at $2$, while maximal unipotent subgroups should increase their nilpotency class with their rank. Maybe $\dim(V)=3$ is small enough though.

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This looks a lot like the definition of the Heisenberg group on a general symplectic vector space. (See Wikipedia.) Maybe it's related? –  Jim Belk Feb 12 '11 at 8:28
    
I suspect it is some sort of amalgamated product of such things, but I don't know how to use that. The Heisenberg groups have dimension 2n+1, but this guy has dimension n(n+1)/2. In particular, a Heisenberg group has odd dimension, but when n=3, wedgey has dimension 6. –  Jack Schmidt Feb 12 '11 at 15:25
    
@Jack: To me, at least for characteristic different from $2$, it looks like you are taking some variant of the $2$-nilpotent product of $\mathbb{F}^n$ with itself (where $n$ is the dimension); that is, just the $2$-nilpotent product of the underlying abelian groups of $V$. Haven't checked it through, but if $2$ is invertible, then every pure wedge can be obtained as the commutator of two elements, and the multiplication rule you give is almost the multiplication rule in the $2$-nilpotent product (except you have the square root of the commutator instead of the commutator). –  Arturo Magidin Feb 12 '11 at 21:02
    
@Arturo: Awesome, thanks! That improves my "some sort" very clearly. The factor on the wedge can be changed (by an invertible factor) without altering the isomorphism type, at least in dim(V)=2 case, and I think in general. So I've got the nilpotent product of V with itself. That'll help, but I'm not sure if nilpotent products have good representation properties. If you find anything about how to write nilpotent products as matrices, that'll be a perfect answer. I find it hard to believe it doesn't look something like the Heisenberg group. –  Jack Schmidt Feb 12 '11 at 22:48
    
@Jack: Well, the Heisenberg group is exactly a $2$-nilpotent product (over $\mathbb{Z}_n$, it's the $2$-nilpotent product of $\mathbb{Z}_n$ with itself, for example). In general, if $G$ and $H$ are groups, their two nilpotent product consists of all elements of the form $gh\alpha$, with $g\in G$, $h\in H$, $\alpha\in H^{\rm ab}\otimes G^{\rm ab}$, with product given by $$(g_1h_1\alpha)(g_2h_2\beta) = (g_1g_2)(h_1h_2)(h_1\otimes g_2 + \alpha+\beta)$$so you can see how it looks similar to me... –  Arturo Magidin Feb 12 '11 at 23:08

1 Answer 1

up vote 3 down vote accepted

I think you have a nil-2 product of $n$ copies of $F$ (rather than two copies of $V$, as I originally wrote in comments).

If $A_1,\ldots,A_n$ are abelian groups, then their 2-nilpotent product is their free product modulo the second term of the lower central series of the free product (in general, you take the free product and mod out by the intersection of the second term of the lower central series with the cartesian, the kernel of the map from the free to the direct product).

For the abelian case, the elements of the 2-nil product can be written uniquely as $$a_1\cdots a_n\gamma$$ where $a_i\in A_i$, and $\gamma$ is in the commutator subgroup. The commutator subgroup is isomorphic to $$\sum_{1\leq i\lt j\leq n} A_j\otimes A_i.$$ (In the general case, you take the abelianizations).

The multiplication rule is $$(a_1\cdots a_n\gamma)(b_1\cdots b_n\delta) = (a_1b_1\cdots a_nb_n)\left(\gamma+\delta+\sum_{1\leq i\lt j\leq n}b_i\otimes a_j\right).$$

If we take each $A_i$ isomorphic to $F$, then the commutator subgroup will be isomorphic to an $F$-vector space of dimension $\binom{n}{2}$, with basis given by $v_{ji}$, where $1\leq i\lt j\leq n$ (identifying $1_j\otimes 1_i$ with $v_{ji}$). If you make the "convention" that $v_{ij} = -v_{ji}$ and $v_{ii}=0$, then you get exactly the wedge $F^n\wedge F^n$.

I am a little worried about that factor of $2$ in the commutator; there would be no problem with defining the $2$-nil product of groups of exponent $2$ and getting a 2-nil group (of exponent 4), which you cannot do with your set up. Indeed, if I think of this as the $2$-nilpotent product of $n$-copies of $F$, with elements expressed as sums of multiples of $\mathbf{1}_1,\ldots,\mathbf{1}_n$, I would write: \begin{align*} [\alpha_1\mathbf{1}_1+\cdots+\alpha_n\mathbf{1}_n,\beta_1\mathbf{1}_1+\cdots+\beta_n\mathbf{1}_n] &= \prod_{i,j=1}^n [\alpha_i\mathbf{1}_i,\beta_j\mathbf{1}_j]\\ &= \prod_{1\leq i\lt j\leq n} [\mathbf{1}_j,\mathbf{1}_i]^{\alpha_j\beta_i-\alpha_i\beta_j}. \end{align*}

On the other hand, you would have \begin{align*} 2(\alpha_1\mathbf{v}_1+\cdots+\alpha_n\mathbf{v}_n)\wedge(\beta_1\mathbf{v}_1 + \cdots + \beta_n\mathbf{v}_n) &= 2\sum_{i,j=1}^n \alpha_i\mathbf{v}_i\wedge\beta_j\mathbf{v}_j\\ &= 2\sum_{i,j=1}^n (\alpha_i\beta_j)\mathbf{v}_i\wedge\mathbf{v}_j\\ &= 2\sum_{1\leq i\lt j\leq n} (\alpha_j\beta_i - \beta_j\alpha_i)\mathbf{v}_i\wedge\mathbf{v}_j, \end{align*} that is, twice as much as I do.

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Thanks, this is very nice! I will renew my search for matrix representations of 2-nilpotent products, though I am more hopeful now that the guys being multiplied are just copies of F. The factor of 2 can be fixed without altering the isomorphism type (when char(F)≠2 of course). I also checked a few specific F and n in GAP, and it looks good. –  Jack Schmidt Feb 13 '11 at 5:11
    
Thanks, this is looking very good. When F is a prime field (or Z if one is feeling particularly loose in the definition of field), then these are just free nilpotent groups of class 2 and rank dim(V), and apparently they have really quite good matrix representations (independent of characteristic). –  Jack Schmidt Feb 13 '11 at 5:48
    
@Jack. Yes, the 2-nil product of n copies of Z (resp. $\mathbf{F}_p$) is the relatively free nilpotent group of class two, rank $n$ (resp. exponent $p$). I'm glad it's looking useful. –  Arturo Magidin Feb 13 '11 at 5:50

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