We know that for $0\leq k\leq n$, the $k$-th de Rham cohomology group of the $n$-torus $\mathbb{T}^n = S^1 \times \dots \times S^1$ is $$H^k_{dR}(\mathbb{T}^n) \cong \mathbb{R}^{n\choose k}.$$
I just saw a theorem:
Theorem (Cartan-Eilenberg). For a connected compact Lie group $G$, its de Rham cohomology is isomorphic to the cohomology of the subcomplex of left-invariant differential forms on $G$.
Now I am trying to understand the following example (on page 138 of Morita's "Geometry of Differential Forms"):
Since the n-dimensional torus $\mathbb{T}^n$ can be regarded as the $n$-fold product of $SO(2)$, it is a connected compact Lie group. Its Lie algebra is commutative and can be naturally identified with $\mathbb{R^n}$. Therefore, by the theorem, we have $$ H^k_{dR}(\mathbb{T}) \cong \bigwedge^k \mathbb{R^n} = \mathbb{R}^{n\choose k}.$$
This is confusing to me: when the Lie algebra is $\mathbb{R}^n$, the subcomplex of left-invariant differential forms on $\mathbb{T}^n$ should be the complex of differential forms on $\mathbb{R}^n$, because a left-invariant differential form is determined by its action on the Lie algebra. Then when we compute the cohomology of this complex, we should get the de Rham cohomology of $\mathbb{R}^n$, which is very different from that of $\mathbb{T}^n$.
Could anybody point out my mistakes in the above reasoning? Thank you!
Update: I just realized that to be left-invariant, the differential $k$-form must be an $\mathbb{R}$-linear combination of $dx_{i_1}\wedge\dots\wedge dx_{i_k}$. So, forms like $f(x) dx_{i_1}\wedge\dots\wedge dx_{i_k}$ is not allowed if $f$ is non-constant. Then, everything is in the kernel while only $0$ is in the image. Hence the result follows.