Return to Answer

2-step nilpotent Lie algebra structures on a finite-dimensional vector space $L$ of dimension $3 \mid n$ over $\mathbb{F}_q$.

2-step nilpotent Lie brackets on $\mathbb{F}_q^n$ when $3 \mid n$.

To get a lower bound on the number of isomorphism classes we quotient badly by the action of $GL_n(\mathbb{F}_q)$ by dividing by $q^{n^2}$ (it should be possible to be somewhat more careful than this), and we get that there are at least

$$q^{ \frac{2}{27} n^3 - \frac{8n^2}{9} - \frac{n}{3}}$$

isomorphism classes of 2-step nilpotent Lie algebras of dimension $3 \mid n$ over $\mathbb{F}_q$, with similar explicit bounds for $n \equiv 1, 2 \bmod 3$. It is maybe surprising that it's possible to prove a matching upper bound, at least up to leading order in the exponent; I don't know what that argument looks like in detail.

To get a lower bound on the number of isomorphism classes we quotient badly by the action of $GL_n(\mathbb{F}_q)$. At this point we can actually restore the factor of $(q - 1)^b$ we lost above (although it doesn't matter too much either way); it's not hard to show that $\frac{|GL_n(\mathbb{F}_q)|}{|GL_b(\mathbb{F}_q)|} \le q^{n^2 - b^2}$, so we can then divide by $|GL_b(\mathbb{F}_q)|$ and then by $q^{n^2 - b^2}$ to get a lower bound, whereupon what remains is a polynomial in $q$ with non-negative coefficients which can be bounded from below by its leading term again. We get that there are at least

$$q^{ \frac{2}{27} n^3 - \frac{8n^2}{9}}$$

isomorphism classes of 2-step nilpotent Lie algebras of dimension $3 \mid n$ over $\mathbb{F}_q$. It is maybe surprising that it's possible to prove a matching upper bound, at least up to leading order in the exponent; I don't know what that argument looks like in detail.

ways to make the above choices. Now our job is to find $a, b$ which maximizes this, or at least which makes it quite big since we're aiming for a lower bound. The leading term in $q$ (which is a lower bound since all the coefficients are positive!) is $q$ to the power of

Subject to the constraint that $a + b = n$ this is maximized when $a \approx \frac{2n}{3}, b \approx \frac{n}{3}$, and we could be more careful depending on the value of $n \bmod 3$ if desired. Let me instead restrict to the case that $3 \mid n$ so that we can divide by $3$ exactly, so our leading term and lower bound is that there are at least

$$q^{ \frac{2}{27} n^3 + \frac{2n^2}{3} }$$

2-step nilpotent Lie algebra structures on a finite-dimensional vector space $L$ of dimension $3 \mid n$ over $\mathbb{F}_q$. You didn't ask for this, but I want to mention it anyway: quotienting by the action of $GL_n(\mathbb{F}_q)$ to get isomorphism classes at worst divides this by $q^{n^2}$, so we get that there are at least

$$q^{ \frac{2}{27} n^3 - \frac{n^2}{3} }$$

isomorphism classes of 2-step nilpotent Lie algebras of dimension $3 \mid n$ over $\mathbb{F}_q$, with similar explicit bounds for $n \equiv 1, 2 \bmod 3$. It is maybe surprising that it's possible to prove a matching upper bound, at least up to leading order in the exponent; I don't know what that argument looks like in detail. Plugging in some actual numbers, we get that there are

• at least $q^8$ Lie brackets on $\mathbb{F}_q^3$ and so at least $q^{-1}$ isomorphism classes of $3$-dimensional lie algebras (technically true!)
• at least $q^{40}$ Lie brackets on $\mathbb{F}_q^6$ so at least $q^4$ isomorphism classes of $6$-dimensional Lie algebras, and
• at least $q^{108}$ Lie brackets on $\mathbb{F}_q^9$ so at least $q^{27}$ isomorphism classes of $9$-dimensional Lie algebras.

ways to make the above choices. Now our job is to find $a, b$ which maximizes this, or at least which makes it quite big since we're aiming for a lower bound. The leading term in $q$ is $q$ to the power of

Subject to the constraint that $a + b = n$ this is maximized when $a \approx \frac{2n}{3}, b \approx \frac{n}{3}$, and we could be more careful depending on the value of $n \bmod 3$ if desired. Let's instead restrict to the case that $3 \mid n$ so that we can divide by $3$ exactly, and also take the liberty of dividing by $(q - 1)^b$ so that what remains is a polynomial in $q$ with nonnegative coefficients and so the leading term is a true lower bound. We get that there are at least

$$q^{ \frac{2}{27} n^3 + \frac{n^2}{9} - \frac{n}{3}}$$

2-step nilpotent Lie algebra structures on a finite-dimensional vector space $L$ of dimension $3 \mid n$ over $\mathbb{F}_q$.

To get a lower bound on the number of isomorphism classes we quotient badly by the action of $GL_n(\mathbb{F}_q)$ by dividing by $q^{n^2}$ (it should be possible to be somewhat more careful than this), and we get that there are at least

$$q^{ \frac{2}{27} n^3 - \frac{8n^2}{9} - \frac{n}{3}}$$

isomorphism classes of 2-step nilpotent Lie algebras of dimension $3 \mid n$ over $\mathbb{F}_q$, with similar explicit bounds for $n \equiv 1, 2 \bmod 3$. It is maybe surprising that it's possible to prove a matching upper bound, at least up to leading order in the exponent; I don't know what that argument looks like in detail.