# Tagged Questions

**1**

vote

**0**answers

54 views

### Adjoint action of semi-direct product

Let $G$ and $H$ be Lie groups with associated Lie algebras $\mathfrak{g}:=\text{Lie}(G)$ and $\mathfrak{h}:=\text{Lie}(H)$ and adjoint actions
$\text{Ad}^G:G \to \text{Aut}_\text{Lie}(\mathfrak{g})$ ...

**8**

votes

**3**answers

445 views

### What is the categorical significance of the trivial $\mathfrak{g}$-module in the category of $\mathfrak{g}$-mod?

This question may be trivial for experts.
Let $\mathfrak{g}$ be a Lie algebra over a field $k$ and consider the category $\mathfrak{g}$-mod of $\mathfrak{g}$ modules. We can add suitable conditions, ...

**15**

votes

**2**answers

563 views

### Is a retract of a free object free?

I wonder whether this is true in the categories of groups, monoids, commutative algebras, associative algebras, Lie algebras?

**0**

votes

**0**answers

244 views

### cokernel for $L_\infty$-algebra morphisms

As I have asked a wrong question previously, I edited a bit.
It is obvious that the cokernel construction does not work well for the category of Lie algebras. The cokernel exists only for a normal ...

**18**

votes

**3**answers

1k views

### Why the BGG category O?

Given a finite-dimensional semisimple complex Lie algebra $\mathfrak{g}$, the Bernstein-Gelfand-Gelfand category $\mathcal O$ is the full subcategory of $\mathfrak g$-modules satisfying some ...

**3**

votes

**1**answer

211 views

### A functor that comes from a morphism in a bigger category

My loose question is like this: what would you say about an equivalence of categories where both are concrete categories, and the equivalence functor is induced from a set-theoretic bijection at the ...

**34**

votes

**2**answers

1k views

### What algebraic group does Tannaka-Krein reconstruct when fed the category of modules of a non-algebraic Lie algebra?

Let $\mathfrak g$ be a finite-dimensional Lie algebra over $\mathbb C$, and let $\mathfrak g \text{-rep}$ be its category of finite-dimensional modules. Then $\mathfrak g\text{-rep}$ comes equipped ...