Existence of left invariant $n-$forms in a lie Group I am trying to show that the space of left invariant $n$-forms for an $n-$dimensional Lie group $G$ has dimension $1$. Now I was able to see that if we have a left-invariant $1-$form then it's value depends only at the values of $\omega_e$ has so the maximum dimension of this space will be $1$. Now I am trying to prove the existence of such forms. For this take an element $\omega_e \in \Lambda^n(T_eG)$ and define $\omega_g(X_1,...,X_n)=\omega_e((dL_{g^{-1}})_g X_1,...,(dL_{g^{-1}})_gX_n)$, and it's easy to see that is left-invariant. Now my problem is checking that in fact we have smoothness. For this I tried seeing that $\omega$ defines a multilinear alernating map from $\mathfrak{X}(M)\times ... \times \mathfrak{X}(M)\rightarrow C^{\infty}(M)$. Now the problem is that if I have vector fields $X_1,...,X_n$ I can't seem to argument why the function $\omega(X_1,...,X_n)(g)=\omega_e((dL_{g^{-1}})_g X_1(g),...,(dL_{g^{-1}})_g X_n(g) )$ is smooth , since we have that the $L_g^{-1}$ are changing . Any help with this last part is aprecciated . Thanks in advance.
 A: You already proved that a left invariant $n$-form would be unique up to a constant. Now, to prove that there indeed exist such a left invariant $n$-form, let $X_1,\ldots,X_n$ be a basis of $T_eG$. Let $\theta^1,\ldots,\theta^n$ be its dual basis in $T_eG^*$. Then $\omega_e=\theta^1\wedge\cdots\wedge \theta^n$ is a non-zero differential form of degree $n$ in $T_eG$ because $\omega_e(X_1\wedge\cdots\wedge X_n) = 1$. Now, one can define a $n$-form $\omega$ on $G$ by requiring it to be left invariant, that is
$$
\omega_g = \omega_e\left((\mathrm{d}L_{g^{-1}})\cdot,\ldots,\left(\mathrm{d}L_{g^{-1}}\right)\cdot \right)
$$
It is a smooth $n$-form, and it is left invariant by construction. The smoothness follows from the fact that $g \mapsto \mathrm{d}L_{g^{-1}}$ is smooth because $G$ is a Lie group! Then, if $X$ is a vector field, in the left invariant frame $X_1,\ldots,X_n$ created from the basis on $T_eG$, one can write
$$
Y(g) =\sum_{i=1}^n Y^i(g)\left(\mathrm{d}L_gX_i\right)
$$
and then
$$
\omega_g(Y_1(g),\ldots,Y_n(g)) = \omega_g\left(\sum_{i}Y_1^j(g)X_i(g),\cdots,\sum_{i}Y_n^j(g)X_i(g) \right)
$$
And using linearity and left-invariance of $\omega_g$ and $X_j(g)$, the smoothness follows from the smoothness of the function $g\mapsto Y_j^i(g)$. Hence, $\omega$ is smooth.
