截断多项式的Novikov代数的上同调与扩张
Cohomology and extensions of Novikov algebras of truncated polynomials
AI总结：
该研究计算了特征p>0域上截断多项式Novikov代数的二阶上同调，给出不同情形下的维数与显式上闭链，证明特征0对应代数刚性而截断版本非刚性，揭示截断是破坏刚性的关键因素。
AI中文摘要：
设$\bk$是特征为$p>0$的域，不假设其代数闭，令$V=\bk[x]/(x^p)$为乘积为$a\b b=ab'$的Novikov代数。对$\blambda\bk$，令$M(\blambda)$为Xu模。我们计算了所有$\blambda$和$p$对应的二阶上同调$\text{H}^2(V,M(\blambda))$，并描述了相关的阿贝尔扩张。计算利用了$\text{Z}/p\text{Z}$分次表示$V\bk[t]/(t^p-1)$，以规避截断问题并简化上闭链恒等式。我们确定了$\text{H}^2(V,M(\blambda))$的精确维数：当$\blambda\notin\text{F}_p$时维数为0，当$\blambda\text{F}_p$且$p$为奇数时维数为3，当$\blambda\text{F}_p$且$p=2$时维数为4。我们给出了所有情形下的显式上闭链代表元。作为推论，我们证明当$\blambda\notin\text{F}_p$时，$V$通过$M(\blambda)$的每个阿贝尔扩张都是分裂的。我们还处理了特征0的类似对象$P=\bk[t]$：Xu的参数$\blambda$坍缩为单一值$\blambda=0$，且始终有$\text{H}^2(P,M(\blambda))=0$，因此$P$是刚性的（实际上是形式刚性的），而其正特征截断版本则从不具有刚性——在该族中，破坏刚性的是截断而非正特征本身。
英文摘要：
Let $\kk$ be a \emph{field of characteristic $p>0$, not assumed algebraically closed}, and let $V=\kk[x]/(x^p)$ be the Novikov algebra with product $a\circ b=ab'$. For $λ\in\kk$, let $M(λ)$ be Xu's module. We compute the second cohomology $\Ht(V,M(λ))$ for all $λ$ and $p$, and describe the associated abelian extensions. The computation utilizes a $\Zp$-graded presentation $V\cong\kk[t]/(t^p-1)$ to bypass truncation issues and simplify cocycle identities. We determine the exact dimensions of $\Ht(V,M(λ))$, showing it is $0$ for $λ\notin\Fp$, $3$ for $λ\in\Fp$ with odd $p$, and $4$ for $λ\in\Fp$ with $p=2$. Explicit cocycle representatives are provided for all cases. As corollaries, we show that every abelian extension of $V$ by $M(λ)$ splits when $λ\notin\Fp$. We also treat the characteristic-$0$ analogue $P=\kk[t]$: Xu's parameter $λ$ collapses to the single value $λ=0$, and $\Ht(P,M(λ))=0$ throughout, so $P$ is rigid (in fact formally rigid) while its positive-characteristic truncation never is --- within this family, it is truncation rather than positive characteristic per se that destroys rigidity.
