One-one correspondence between 2-Cohomology Group and (Equivalent) Abelian extension of Lie Algebra.
Sorry about any ambiguation, because I'm not familiar with homological algebra and category theory.
And any corrections or clarifications are grateful.
We have Chevalley-Eilenberg Complex, and we can use this complex to calculate cohomology group (or $\mathbb{F}$-vector space) of it.
An application of Chevalley-Eilenberg Complex is about Abelian extension of a given lie algebra.
So here is one-one correspondence between 2-Cohomology Group and (Equivalent) Abelian extension of Lie Algebra.
(I have a question about the proof of the theorem below)
Theorem Let $V$ be a finite-dimensional representation of $\mathfrak{g}$. Regard $V$ as an abelian Lie algebra. Then there exists a bijection between $H^2(\mathfrak{g}, V )$ and the set of equivalence classes of abelian extensions of $\mathfrak{g}$ by $V$.
Remark
1.Set $\mathcal{C}^q=\mathcal{C}^q(\mathfrak{g},V):=\operatorname{Hom}_F(\bigwedge^q\mathfrak{g},V)=(\bigwedge^q\mathfrak{g})^*\otimes V, q\ge 0$, And the Chevalley-Eilenberg Complex is $C^0\xrightarrow[]{d} C^1\xrightarrow[]{d}C^2\xrightarrow[]{d}C^3\xrightarrow[]{d}\dots$, $\mathrm{H}^q(\mathfrak{g},V):=\mathrm{ker}(d:\mathcal{C}^q\to\mathcal{C}^{q+1})/\mathrm{im}(d:\mathcal{C}^{q-1}\to\mathcal{C}^q).$
2.The exact sequence $0\to V\to\mathfrak{h}\to\mathfrak{g}\to0$ is called Abelian extension of $\mathfrak{g}$ by $V$, if all arrows are lie algebra homomorphism, and $V$ is an abelian lie algebra.
3.The equivalent extensions means that, there exist a commutative diagram: $$ \begin{matrix}0&\longrightarrow&\mathfrak{a}&\longrightarrow&\mathfrak{g}&\longrightarrow&\mathfrak{b}&\longrightarrow&0\\ =\downarrow&&=\downarrow&&\varphi\downarrow&&=\downarrow& &=\downarrow\\ 0&\longrightarrow&\mathfrak{a}&\longrightarrow&\mathfrak{h}&\longrightarrow&\mathfrak{b}&\longrightarrow&0\end{matrix} $$ And $\varphi:\mathfrak{g}\to\mathfrak{h}$ is a lie algebra isomorphism.
Let me state the proof briefly.
Step1: prove the Lemma 1
The exact sequence (of lie algebra) $0\rightarrow\mathfrak{a}\rightarrow\mathfrak{g}\rightarrow\mathfrak{b}\rightarrow0$ splits iff $\mathfrak{g}\simeq\mathfrak{b}\ltimes_{\pi}\mathfrak{a}$.
Remark
$\mathfrak{b}\ltimes_{\pi}\mathfrak{a}$ is a lie algebra, with underlying set $\mathfrak{b}\times\mathfrak{a}$ and the lie bracket defined as $$ [(B_1,A_1),(B_2,A_2)]:=([B_1,B_2],\pi(B_1)A_2-\pi(B_2)A_1+[A_1,A_2]),\ A_1,A_2\in\mathfrak{a}, B_1,B_2\in\mathfrak{b} $$
And $\pi:\mathfrak{b}\to \text{Der}_\mathbb{F}\mathfrak\ {a}$.
Step2: prove the Lemma 2
Let $V$ be a finite dimensional representation of $\mathfrak{g}$. Then for any 2-cocycle $\omega$, define a bilinear map on the vector space $\mathfrak{h}:= \mathfrak{g}\oplus V$ . $$ [(X,v),(Y,u)]:=([X,Y],Xu-Yv+\omega(X,Y)). $$ The above bilinear map is indeed a Lie bracket, i.e, it is skew-symmetric and satisfies Jacobi identity. We write $\mathfrak{h}_{\omega}=\mathfrak{g}\oplus_{\omega}V$ for this Lie algebra. Hence $$ 0\rightarrow V\rightarrow\mathfrak h_{\omega}\rightarrow\mathfrak g\rightarrow0 $$ is an Abelian extension of $\mathfrak{g}$ by $V$.
Step3: prove Lemma 3: the converse part of Lemma 2
Suppose that we have an abelian extension of Lie algebra $$ 0\to V\xrightarrow{\iota}\mathfrak{h}\xrightarrow{\pi}\mathfrak{g}\to0. $$ The above exact sequence splits in the category of vector spaces, hence we can choose a section $s:\mathfrak{g}\rightarrow\mathfrak{h}$ (a linear map such that $\pi\circ s=\mathrm{Id}_{\mathfrak{g}}$).
Define a linear map $\rho:\mathfrak{g}\to\operatorname{End}_{F}(V)$ via $\rho_{s}(X)(v):=[s(X),\iota(v)]$. This map is well-defined (i.e. independent of the choice of $s$). Thus, $\rho:\mathfrak{g}\to\operatorname{End}_{F}(V)$ is a representation of $\mathfrak{g}$ on $V$ (which can be checked easily by Jacobi identity). In this way, any abelian extension: $$ 0\to V\xrightarrow{\iota}\mathfrak{h}\xrightarrow{\pi}\mathfrak{g}\to0. $$ naturally provides a $\mathfrak{g}$-module $V$ . Then it makes sense to discuss $\mathrm{~H}^{\ast}(\mathfrak{g},V)$.
We identify $V$ with the subspace $\iota(V)$ in $\mathfrak{g}$. Take an arbitrary section $s:\mathfrak{g}\rightarrow\mathfrak{h}$, Define a function $$ \omega_s:\mathfrak{g}\times\mathfrak{g}\to\mathfrak{h},\ \omega_s(X,Y):=s([X,Y])-[s(X),s(Y)]. $$ Then $\mathrm{Im}\omega_{s}\subseteq\operatorname{ker}\pi=V$ . Thus $\omega_s:\mathfrak{g}\times\mathfrak{g}\to V$. And (Lemma 3) $\omega_s$ is a is a 2-cocycle.
And here is my Question1:
In the Theorem above, $V$ is already a representation of $\mathfrak{g}$, which means the representation map $\mathfrak{g}\to \text{End}_\mathbb{F}V$ is given and fixed.
But in Lemma 3, given an Abelian extension, the exact sequence $0\to V\xrightarrow{\iota}\mathfrak{h}\xrightarrow{\pi}\mathfrak{g}\to0$ itself provides another representation of $V$ on $\mathfrak{g}$. In other word, it may be different from the original one.
And we continue the (outline) of proof.
Step4: prove the Lemma 4
Let $$ 0\to V\xrightarrow{\iota}\mathfrak{h}\xrightarrow{\pi}\mathfrak{g}\to0. $$ be an abelian extension (And this exact sequence also provides a $\mathfrak{g}$-module $V$). Take $s : \mathfrak{g}\to\mathfrak{h}\simeq (\mathfrak{g}\ltimes V), X\mapsto(X, 0)$ and construct $\omega_s$ as in Lemma 3. Then the identity map on the vector space $\mathfrak{g}\ltimes V\simeq \mathfrak{h}_{\omega_s}$ is a Lie algebra isomorphism (And equivalence extension).
Step5: prove the Lemma 5
Let $s,s^{\prime}:\mathfrak{g}\rightarrow\mathfrak{h}$ be two sections. Construct $\omega_{s'}$ and $\omega_s$ as in Lemma 3. Then they are cohomological, i.e. they define the same cohomology class in $H^2(\mathfrak{g}, V )$.
And $\mathfrak{h}_{\omega_s},\mathfrak{h}_{\omega_s'}$ are equivalent extension.
Step6: prove the Lemma 6
Let $\alpha$ be a 2-cocycle. Construct $\mathfrak{h}_\alpha$ as in Lemma 2. Then $\alpha$ can be recovered from the section $s : \mathfrak{g}\to\mathfrak{h}_\alpha, X\mapsto(X, 0)$. In other words, $\alpha(X,Y)=[s(X),s(Y)]-s([X,Y]).$
Step7:
Lemma 2 says that we can construct an abelian extension $\mathfrak{h}_\alpha$ out of a 2-cocycle $\alpha$ .
Lemma 3 we can construct 2-cocycles from any abelian extension (such as $\mathfrak{h}_\alpha$).
Lemma 4 and Lemma 6 implies that these two process are inverse to each other.
Lemma 5 The construction of 2-cocycle from Lie algebra extension depends on the choice of the section $s : \mathfrak{g}\to\mathfrak{h}$. Yet two different choices of the section yields the same cohomology class (Note in Lemma 3 the representation of $\mathfrak{g} $ on $V$ is induced by abelian extension) as well as the same equivalence class of the extension. This finishes the proof of the Theorem.
(I'm confused with this proof, mainly because the Question1 above)
And I think because in Lemma 3 we can't fix the representation of $\mathfrak{g}$ on $V$, so the one-one correspondence should be $$ \{\text{All abelian extension of}\ \mathfrak{g}\ \text{by}\ V\}\\ \stackrel{\mathrm{1:1}}{\longleftrightarrow}\\ \bigsqcup_{\text{All representation of $\mathfrak{g}$ on $V$, each of them induce one Chevalley-Eilenberg Complex}}\text{H}^2(\mathfrak{g},V_{\text{fixed representation}}) $$ But I think it is not the right answer either, because there is another question below.
Question2 There is an example
Consider Abelian extension of $\mathbb{R}^2$(as abelian lie algebra) by $\mathbb{R}$.
There will be a split extension $0\to\mathbb{R}\to\mathbb{R}^3\to\mathbb{R}^2\to0$ ($\mathbb{R}^3$ is abelian lie algebra), and a unique non-split extension $0\to\mathbb{R}\to\mathfrak{h}_1\to\mathbb{R}^2\to0$, here $\mathfrak{h}_1$ is Heisenberg Lie algebra.
But $\text{H}^2(\mathbb{R}^2, \mathbb{R}) = \mathbb{R}$ (Regard $\mathbb{R}$ as a trivial $\mathbb{R}^2$-module), which means there are infinitely many ($\text{card}(\mathbb{R})$) equivalence class of abelian extension of of $\mathbb{R}^2$ by $\mathbb{R}$. But we have shown there are only 2 equivalence class.
Thank you for finishing reading this super long and verbose question.
And thanks in advance if you can give me some hint, correction or reference!
I watched http://math.stanford.edu/~conrad/210BPage/handouts/Cohomology&Extensions.pdf today, and all questions are solved.
To Question1.
The action of $\mathfrak{g}$ on $V$ must be a fixed one.
Which means, we need to state the 1:1 correspondence theorem as below:
$$ \{\text{Abelian Extension inducing the given}\ \mathfrak{g}-\text{module}\ V\} \leftrightarrow H^2(\mathfrak{g},V) $$
To Question2.
There are some automorphism $\varphi$ of Heisenberg Lie Algebra itself, that cannot be inserted into the (commutative) diagram (of equivalence Abelian Extension).
Regrading the underlying set of Heisenberg Lie Algebra $\mathfrak{h}_1$ as $(\mathfrak{g}=\mathbb{R}^2)\times\mathbb{R}$.
Since $\mathfrak{g}$ acts on $\mathbb{R}$ trivially, so the lie bracket is defined as.
$$ [(X,a),(Y,b)] = ([X,Y],\omega(X,Y)) $$
All nonzero 2-cocycle are linearly dependent ($H^2(\mathbb{R}^2,\mathbb{R})=\mathbb{R}$), so the lie brackets in all generated Lie Algebra $\mathfrak{h}_1^r$ should be.
$$ \text{in}\ \mathfrak{h}_1^r:\ [(X,a),(Y,b)] = ([X,Y],r\omega(X,Y)),\ r\in\mathbb{R}\setminus\{0\} $$
All $\mathfrak{h}_1^r$ are isomorphic as Lie Algrbra.
Because we have $\varphi_r:\mathfrak{h}_1\to\mathfrak{h}_1^r $
$$ \varphi_r((X,a)) = (X, ra) $$
But the following diagram is not commute (for the 2nd cube from the left).
$$ \begin{matrix}0&\longrightarrow&\mathbb{R}&\longrightarrow&\mathfrak{h_1}&\longrightarrow&\mathbb{R}^2&\longrightarrow&0\\ =\downarrow&&=\downarrow&&\varphi_r\downarrow&&=\downarrow&&=\downarrow\\ 0&\longrightarrow&\mathbb{R}&\longrightarrow&\mathfrak{h}_1^r&\longrightarrow&\mathbb{R}^2&\longrightarrow&0\end{matrix} $$