Sofic Lie Algebras

Clearly, we can embed $\prod_{k\to\omega}\mathfrak{sl}_{n_{k}}(F)/\ker\rho_{\omega}\hookrightarrow\prod_{k\to\omega}\mathfrak{gl}_{n_{k}}(F)/\ker\rho_{\omega}$. Moreover, for any sequence $(A_{k})\in\prod_{k\to\omega}\mathfrak{gl}_{n_{k}}(F)/\ker\rho_{\omega}$, we can consider the second sequence $(A_{k}-E_{11}\text{tr}(A))$. Then

so $(A_{k})=(A_{k}-E_{11}\text{tr}(A))$ in the ultra-product. Thus the natural embedding is a surjection. ∎

If $\omega$ is a principal ultrafilter, this is trivially true. So we assume that $\omega$ is free.

Denote $R:=\prod_{k\to\omega}M_{n_{k}}(F)/\ker\rho_{\omega}$. Suppose $0\neq x\in R$. Let $\delta:=\rho_{\omega}(x)$. Since $\delta>0$, there exists $n\in\mathbb{N}$ such that $\frac{1}{n}<\delta$. Thus, if $(B_{k})\in\prod M_{n_{k}}(F)$ is a representative for $x$, there exists some $S\in\omega$ such that $\rho_{n_{k}}(B_{k})\geq\frac{1}{n}$ for every $k\in S$.

Let $J_{k}$ be the Jordan normal form of $B_{k}$. Then $J_{k}$ row equivalent to

where $0\leq m\leq n-1$. Thus, there exists $P_{k},Q_{k}\in M_{n_{k}}(F)$ such that

Define $C_{k}\in M_{n_{k}}(F)$ via

Then if $y=[C_{k}]\in R$, we have that $y\in(x)$, the ideal generated by $x$. Moreover, we notice that for $k\in S$,

Let $\varepsilon>0$. Since $\omega$ is a free ultrafilter, $S$ must be infinite. Therefore, there exists an infinite subset $S^{\prime}\subset S$ such that $\rho_{n_{k}}(C_{k}-I_{n_{k}})<\varepsilon$. Notice that $S^{\prime}\in\omega$ since otherwise, $(S^{\prime})^{C}\cap S\in\omega$ making $\omega$ a principal ultrafilter. Therefore $\rho_{\omega}(y-[I_{n_{k}}])=0$ and $(x)=R$. ∎

This follows from Theorem 2 of [Her61]. ∎

Suppose $p\in SR(L)$ and let $\Theta:L\to\prod_{\omega}\mathfrak{gl}_{n_{k}}(F)/\rho_{\omega}$ be a Lie algebra homomorphism with lifts $\theta_{k}:L\to\mathfrak{gl}_{n_{k}}(F)$. Fix $\delta>0$ and choose $n_{\delta}$ and finite dimensional $W\subset L$ from the definition of the sofic radical. Choose $0<\varepsilon<n_{\delta}$. Then there exists $S\in\omega$ such that $\theta_{k}|_{W}$ is an $\varepsilon$-almost representation for $W$ for every $k\in S$. Therefore $\dim\text{Im }\theta_{k}(p)<\delta n_{k}$. In other words, $\rho_{n_{k}}(\theta_{k}(p))<\delta$ for every $k\in H$. Therefore $\rho_{\omega}(\Theta(p))<\delta$. Since $\delta$ was arbitrary, we get that $\Theta(p)=0$.

Now suppose that $p\in L\setminus SR(L)$. Then there exists $\delta>0$ such that for any finite dimensional $W\subset L$ containing $p$ and $n>0$, there exists $0<\varepsilon<n$ and $\varepsilon$-almost representation $\varphi:L\to\mathfrak{gl}(V)$ such that $\dim\text{Im }\varphi(p)\geq\delta\cdot\dim V$.

Let $W_{k}$ be an increasing sequence of finite dimensional subspaces of $L$ containing $p$ that cover $L$. Then there exists a sequence of $\varepsilon_{k}$ that converges to 0 and $\varepsilon_{k}$-almost representations $\theta_{k}:W_{k}\to\mathfrak{gl}(V_{k})$ such that $\dim\text{Im }\theta_{k}(p)\geq\delta\cdot\dim V_{k}$. Define a map $\hat{\Theta}:L\to\prod\mathfrak{gl}(V_{k})$ by $\hat{\Theta}(x)=(\hat{\theta}_{k}(x))$ where

Choose a non-principal ultrafilter $\omega$ of $\mathbb{N}$ and let $\Theta:L\to\prod_{\omega}\mathfrak{gl}(V_{k})/\rho_{\omega}$ be the composition of $\hat{\Theta}$ with the quotient map. Then since $\varepsilon_{k}\to 0$, we have that $\Theta$ is a Lie algebra homomorphism and $\rho_{\omega}(\Theta(p))\geq\delta>0$. ∎

The lemma show that

where

so it is clear that $SR(L)$ is an ideal.

Now suppose $0\neq p\in L/SR(L)$ and let $q\in L$ be a pre-image of $p$. Then $q\notin SR(L)$ so there exists a Lie algebra homomorphism $\Theta:L\to\prod_{\omega}\mathfrak{gl}_{n_{k}}(F)/\rho_{\omega}$ such that $\Theta(q)\neq 0$. Since $SR(L)\subset\ker\Theta$, we can get a map $\hat{\Theta}$ by composing $\Theta$ with the quotient $L\to L/SR(L)$. Then we have that $\hat{\Theta}(p)=\Theta(q)\neq 0$ so $p\notin SR(L/SR(L))$. ∎

The forward implication is trivial so we only show the reverse implication.

Let $L$ be a Lie algebra such that $SR(L)=0$. Then for every $p\in L\setminus(0)$, there exists a Lie algebra homomorphism $\Theta_{p}:L\to\prod_{\omega}\mathfrak{gl}_{n_{k,p}}(F)/\rho_{\omega}$ such that $\Theta_{p}(p)\neq 0$.

Let $\{x_{i}\}_{i\in\mathbb{N}}\subset L$ be a basis for $L$ as a vector space. We shall construct maps $\Psi_{m}:L\to\prod_{\omega}\mathfrak{gl}_{n_{k,m}}/\rho_{\omega}$ such that $\ker\Psi_{m}\cap\text{span}\{x_{1},\dots,x_{m}\}=(0)$ for every $m\in\mathbb{N}$.

Let $\Psi_{1}=\Theta_{x_{1}}$. Now suppose for $m\in\mathbb{N}_{\geq 2}$, we have a map $\Psi_{m-1}$ as above. Then

If the dimension is 0, let $\Psi_{m}=\Psi_{m-1}$. Otherwise, choose a non-zero element $y_{m}$ in the intersection. In this case, we define $\Psi_{m}=\Psi_{m-1}\oplus\Theta_{y_{m}}$. Then if $z\in\ker\Psi_{m}\cap\text{span}\{x_{1},\dots,x_{m}\}$, we have that $z\in\ker\Psi_{m-1}\cap\text{span}\{x_{1},\dots,x_{m}\}$. Thus $z=\alpha y_{m}$ for some $\alpha\in F$. However, $z\in\ker\Theta_{y_{m}}$ so $z=0$.

Now we construct a map $\Phi:L\to\prod_{\omega}\mathfrak{gl}_{n_{k}}(F)/\rho_{\omega}$ such that $\ker\Phi\subset\bigcap\ker\Psi_{m}$. Let $n_{k}=n_{k,1}\cdots n_{k,k}$. Tensor each component of $\Psi_{m}$ with an appropriately sized identity matrix gives us maps $\hat{\Psi}_{m}:L\to\prod_{\omega}\mathfrak{gl}_{\hat{n}_{k,m}}(F)/\rho_{\omega}$ where $\hat{n}_{k,m}=n_{k}$ if $m\leq k$ and $n_{k,m}$ otherwise. Let $\hat{\psi}_{k,m}:L\to\mathfrak{gl}_{n_{k}}(F)$ be a lift of $\hat{\Psi}_{m}$ for $m\leq k$.

Define a map $\varphi_{k}:A\to\mathfrak{gl}_{2^{k}n_{k}}$ by

Let $\Phi=\prod_{\omega}\varphi_{k}/\rho_{\omega}$.

A direct calculation gives us that for any $x\in L$

where $\psi_{k,m}:L\to\mathfrak{gl}_{n_{k,m}}$ is a lift of $\Psi_{m}$.

Therefore $\ker\Phi\subset\bigcap\ker\Psi_{m}=(0)$, so we have that $\Phi$ is injective. ∎

Suppose $L$ is an abelian Lie algebra over a field $F$ and suppose that $0\neq p\in L$. Then for any finite dimensional subspace $p\in W\subset L$ and $\varepsilon>0$, there exists an $\varepsilon$-almost representation $\varphi:W\to F\cong\mathfrak{gl}_{1}(F)$ such that $\varphi(p)=1$ and $\varphi(W\setminus Fp)=0$. Thus $\dim\text{Im}\varphi(p)=1=\dim F$ so $p\notin SR(L)$. ∎

Let $L$ be a Lie algebra of subeponential growth generated by the set $X\subset L$. Let $V_{n}$ denote the words in $L$ of length at most $n$. We inductively create a basis for $L$ as follows. Let $\{x_{1},\dots,x_{\dim V_{1}}\}$ be a basis for $V_{1}$. For $n\geq 2$, let $\{x_{\dim V_{n-1}+1},\dots,x_{\dim V_{n}}\}$ be the preimage in $L$ of a basis for $V_{n}/V_{n-1}$. We also define $W_{n}$ to be the subspace of $U(L)$ spanned by words of length at most $n$ and $\gamma(n)=\dim W_{n}$.

For $m>n$, we define a linear map $\varphi_{n,m}:V_{n}\to\mathfrak{gl}(W_{m})$ where

We notice on that for $v\in W_{m-n}$ and $x,y\in V_{n}$,

Let us show that for a fixed $n$ that

Since $L$ is of subexponential growth, we have that $\gamma$ is a function of subexponential growth from [Smi76]. To see this suppose that $f:\mathbb{R}\to\mathbb{R}$ is a function of subexponential growth and $d>0$. We have that if

Then there exists $x_{0}\in\mathbb{R}$ such that

Therefore $f(x)\geq L^{\frac{x-x_{0}}{d-1}}f(x_{0})$ and is therefore not of subexponential growth.

Thus for any $n\in\mathbb{N}$ and $\varepsilon>0$, we can find $\varphi_{n,m}$ such that

Now suppose $p\in L\setminus\{0\}$, $V\subset L$ is a finite dimensional subspace containing $p$, and $\varepsilon>0$. Then there exists $n\in\mathbb{N}$ such that $V\subset V_{n}$. Thus for a sufficiently large $m\in\mathbb{N}$, we have an $\varepsilon$-almost representation $\psi:=\varphi_{n,m}|_{V}$ for $V$. On $W_{m-n}$, we have that $\psi(p)$ works like left multiplication by $p$, considered as an element of $U(L)$. Thus $\dim\text{Im}(\psi(p))\geq\dim W_{m-n}$. By our choice of $m$ to make $\psi$ an $\varepsilon$-almost representation, we have that

Hence the normalized rank of $\psi(p)$ is greater than $1-\varepsilon$.

Thus for any non-zero $p\in L$, any finite dimensional subspace $p\in V\subset L$, and $\varepsilon>0$, we can find an $\varepsilon$-almost representation $\psi$ for $V$ such that the normalized rank of $\psi(p)$ is at least 1/2. Specifically, we can choose $\psi$ to be a $\min\{\varepsilon,1/2\}$-almost representation. Therefore $p\notin SR(L)$ so $SR(L)=(0)$ and $L$ is sofic. ∎

Since $L$ embeds in $U(L)$ the reverse direction is trivial. Thus we only prove the forward direction.

Since $L$ is sofic, there exists a Lie algebra homomorphism

Let $\Theta_{1}=\Theta$ and

for $j\geq 2$. Define a new Lie algebra homomorphism $\Theta^{i}:L\to\bigotimes_{j=1}^{i}A$ by $\Theta^{0}=0$, $\Theta^{1}=\Theta$, and $\Theta^{i}=\Theta^{i-1}\otimes 1+1\otimes\Theta_{i-1}$. Since finite tensor product of matrix rings are again matrix rings, we use the proposition to get a unital algebra homomorphism

such that

We claim that $\psi$ is injective.

Let $\{x_{j}\}_{j\in J}\subset L$ be a basis. First suppose that $x_{j_{1}}\cdots x_{j_{d}}\in\ker\psi$ is a monomial for $d\geq 2$. Then by construction

for all $i\geq 1$. Notice that $\Theta^{d}(x_{j_{1}}\cdots x_{j_{d}})$ will consist of a linear sum of pure tensors of $1$ and products of the $\Theta(x_{j_{i}})$’s. In fact, we have two sets of pure tensors: one set consisting of those with at least one 1 and at least one degree 2 or greater monomial and the other set consisting of pure tensors with only linear monomials and no 1’s, i.e. pure tensors of the form $x_{j_{\sigma(1)}}\otimes\cdots\otimes x_{j_{\sigma(d)}}$ for some $\sigma\in S_{d}$. In fact we get exactly one such pure tensor for every $\sigma\in S_{d}$. Call the first set the lower order tensors and the second set the high order tensors.

Now notice that $\widetilde{\Theta}^{d-1}(x_{j_{1}}\cdots x_{j_{d}})$ is a sum of low order tensors of $\widetilde{\Theta}^{d}(x_{j_{1}}\cdots x_{j_{d}})$ with one of the 1’s removed. Thus, viewing $0$ as $0\otimes 0\otimes\cdots\otimes 0$ ($d-1$ times), we can tensor the equation

by 1 in $d$ spots and adding up all the new equations shows that the low order tensors of $\widetilde{\Theta}^{d}(x_{j_{1}}\cdots x_{j_{d}})$ sum to 0. Therefore, we get that

Now suppose that $i>d$. Then we can see that in the sum of pure tensors in $\widetilde{\Theta}^{i}(x_{j_{1}}\cdots x_{j_{d}})$, every pure tensor has at least one 1 in it. Thus repeating the same process of the tensor the equation $\widetilde{\Theta}^{i-1}(x_{j_{1}}\cdots x_{j_{d}})=0$ with $1$ in $i$ spots and summing up, we get back to the equation $\widetilde{\Theta}^{i}(x_{j_{1}}\cdots x_{j_{d}})=0$.

Now we are ready to prove injectivity of $\psi$. Suppose that $a\in\ker\psi\setminus\{0\}$. Put an order on the indexing set $J$ of the basis for $L$. Then by the Poincaré-Birkoff-Witt Theorem, we can uniquely write $a$ as

where $\alpha_{i}\in F$, $\alpha_{i}\neq 0$ for $i\geq 1$, $e^{(i)}_{j}\geq 1$ for every $i,j$, $d_{i}\geq 1$, and $j^{(i)}_{k}<j^{(i)}_{k+1}$. By applying $\widetilde{\Theta}^{0}$, we get rid of every term but $\alpha_{0}$, so we see that $\alpha_{0}=0$. Let

be the degree of the $i$-th monomial and let $D=\max\{D_{i}\}_{1\leq i\leq n}$. Since we know that $\Theta$ is injective on $L$, we get a contradiction if $D=1$. Thus assume that $D\geq 2$.

Repeating the same process we did on monomials, take the equation $\widetilde{\Theta}^{D-1}(a)=0$, tensor it with 1 in the $D$ possible spots and add up all the equations. Then we subtract that sum from the equation $\widetilde{\Theta}^{D}(a)=0$. For every $i$ such that $D_{i}<D$, the terms coming from the $i$-th monomial will vanish in the new equation. Thus we are on left with the monomials of degree exactly $D$. The equation we are left with is a linear combination of pure tensors of the form $\Theta(x_{j_{1}})\otimes\cdots\otimes\Theta(x_{j_{D}})$. Since our monomials are ordered, the only way to have such pure tensors are equal is if they come from the same original monomial. Therefore we have a non-trivial linear combination. However, $\{\Theta(x_{j})\}_{j\in J}\subset A$ is linearly independent since $\Theta$ is injective, so all the weights in the linear combination must be 0. Since $F$ is characteristic, we get that $\alpha_{i}=0$ for every $i$ such that $D_{i}=D$. This is a contradiction so no such $a$ exists and $\ker\psi=(0)$. ∎