understanding this lie algebra question-Any guide please Definition: Given a Lie algebra $L$ and a field $F$, a Lie module of $L$ is a couple $(M,f)$, where $M$ is a vector space and
$f: L\times M\to M$ which satisfies:

*

*$f$ is linear in each variable.


*$f$ satisfies:
$f([x,y],m)=f(x,f(y,m))-f(y,f(x,m))$.
Question: Let L be a Lie algebra and V a vector space. Let $f : L × V → V$ be a
bilinear map. Define an antisymmetric bilinear operation on $L ⊕ V$ by
the following properties.

*

*Its restriction to $L$ is the bracket.

*Its restriction to $V$ is zero.

*Its value on a pair $(x, v)$ with $x ∈ L$, $v ∈ V$ , is $f(x, v)$.

Prove that
$L ⊕ V$ is a Lie algebra with respect to the operation defined above if
and only if $f$ describes on $V$ a structure of L-module.
I do understand all the definitions I need but don't realize how to apply them here.
When they say restriction to $L$ (respectively $V$) does it mean $f:L\times L\to V$ or $f:L\to V$?
 A: The "it" in "Its restriction ..." refers to that new bilinear operation
$$(L\oplus V) \times (L\oplus V) \rightarrow L\oplus V$$
which has not been given a name. Maybe call it $\lbrace \cdot,\cdot \rbrace$.
Then "restriction" is actually still a bit ambiguous, but they must mean
$\lbrace (l_1,0) , (l_2,0) \rbrace = ([l_1, l_2],0)$
by 1, and
$\lbrace (0,v_1) , (0,v_2) \rbrace = (0,0)$
by 2, and
$\lbrace (l,0) , (0,v) \rbrace = (0, f(l,v))$
by 3.

Added in response to comment:
An antisymmetric bilinear $\{ \cdot, \cdot \}$ satisfying the above conditions is necessarily given on general elements by
$$\{ (l_1, v_1), (l_2, v_2)\} = ([l_1,l_2], f(l_1,v_2)-f(l_2,v_1) )$$
Now for one direction, you just check by hand that this satisfies the Jacobi identity
$$\{\{ (l_1, v_1), (l_2, v_2)\}, (l_3, v_3)\} + \{\{ (l_2, v_2), (l_3, v_3)\}, (l_1, v_1)\} + \{\{ (l_3, v_3), (l_1, v_1)\}, (l_2, v_2)\} =0$$
as soon as the two conditions on the map $f$ are satisfied. Of course you can just focus on the second components because we assume the Lie bracket $[, ]$ on $L$ satisfies Jacobi anyway. It's still a bit cumbersome, in total you spell out 12 terms there and see each one is cancelled out by another.
Conversely, to see that $f$ must be as described, assume that $\{ \cdot , \cdot \}$ does satisfy Jacobi, in particular
$$\{\{(l_1,0),(l_2,0)\},(0,v)\} + \{\{(l_2,0),(0,v)\},(l_1,0)\} + \{\{(0,v),(l_1,0)\},(l_2,0)\} =0 $$
and you just check what this means in the second component when written out.
