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If $A$ has positive density in a finite abelian group, then $A+A-A-A$ contains a structured neighbourhood of zero. In a finite-dimensional vector space this neighbourhood can be taken to be a large vector subspace; in a cyclic group it can be taken to be a Bohr set.

A closed vector subspace of a topological vector space is both a vector subspace and a closed set. The kernel of a continuous linear map is a closed vector subspace.

If $A\subseteq {\mathbb{F}}_p^n$ and $|A+A|\le K|A|$, then $A$ is contained in a vector subspace $H$ satisfying

A nonzero Lie algebra representation is irreducible when it has no proper nonzero invariant subspace.

A left ideal $L$ of an associative algebra $A$ is a vector subspace satisfying $AL\subseteq L$. Equivalently, it is a submodule of the left regular $A$-module.

The Freiman-Ruzsa theorem over a finite field states that if $A\subseteq {\mathbb{F}}_p^n$ and $|A+A|\le K|A|$, then $A$ is contained in a vector subspace $H$ withAfter translating $A$, assume $0\in A$. Put $S=A-A$, and choose $X\subseteq A+S$ maximal subject to the translates $x+A$, $x\in X$, being pairwise disjoint. Since $X+A\subseteq 2A+S=3A-A$, the Plünnecke-Ruzsa inequality givesso $|X|\le K^4$.

Maximality gives $A+S\subseteq X+S$: if $y\in A+S$ is not already in $X$, then $(y+A)\cap (x+A)\ne \varnothing$ for some $x\in X$, whence $y\in x+A-A=x+S$. Inductively, $mA+S\subseteq ⟨X⟩+S$ for every positive integer $m$. Because $0\in A$, every element of $⟨A⟩$ belongs to some $mA$ in the finite vector space, and thereforeFinally, $|⟨X⟩|\le p^{|X|}$ and $|S|\le K^2|A|$ by the Plünnecke-Ruzsa inequality, so

Solved by gpt-5.6-sol high.

The hypothesis says that the normalized additive energy of $A$ is at least $\eta$. By the Balog-Szemerédi-Gowers theorem, for an absolute $C_0$ there is $A^{\prime }\subseteq A$ such thatThe finite-field Bogolyubov-Ruzsa consequence of the Freiman-Ruzsa theorem over a finite field says that a set of doubling at most $K$ has a vector subspacewith $|V|\ge p^{-K^{C_1}}|A^{\prime }|$ for an absolute $C_1$. Taking $K={\eta }^{-C_0}$ and enlarging the absolute exponent givesThis is an energy form of the Bogolyubov lemma: the usual lemma assumes positive density in an ambient group, whereas the Balog-Szemerédi-Gowers theorem first extracts a dense structured model from the many additive quadruples. The resulting bound depends on the energy parameter rather than on the possibly tiny ambient density of $A$.

Solved by gpt-5.6-sol high.

Use the Polynomial representation of the Heisenberg Lie algebra on the infinite-dimensional polynomial ring $\mathbb{C}[x]$:The product rule gives $[d/dx,x]=1$, so this is a Lie algebra representation. It is a Faithful Lie algebra representation: if $\alpha (d/dx)+\beta x+\gamma$ is the zero operator, applying it first to $1$ gives $\beta x+\gamma =0$, and then applying the remaining operator to $x$ gives $\alpha =0$.

To prove irreducibility, let $W$ be a nonzero invariant subspace and choose a nonzero polynomial in $W$ of least degree. If its degree were positive, repeated differentiation would produce a nonzero element of smaller degree, so $W$ contains a nonzero constant. Invariance under multiplication by $x$ then puts every monomial $x^n$ in $W$, and hence $W=\mathbb{C}[x]$.

Solved by gpt-5.6-sol high.