I am reading a criterion of Professor Serre for the vanishing of cohomology of Lie algebras in his paper "Sur les groupes de congruence des variétés abéliennes II". To prove this criterion, he used Lemma 1 where I got stuck. Let me state my question shortly.
Let $\mathfrak{g}$ be a sub-Lie algebra of the Lie algebra $\mathfrak{gl}(V)$ where $V$ is a finite-dimensional vector space over a field $K$, and $x \in \mathfrak g$. Denote by $L$ the spectral of $x$. Let $C^n$ be the set of $n$-cochains, i.e., $$C^n=\{ \text{multilinear, alternating maps } \mathfrak g\times \mathfrak g\times...\times \mathfrak g\to V\}.$$ There is a natural action of $x$ on $C^n$, called the Lie derivative, defined by $$(\theta_x f)(y_1,...,y_n)=x\cdot f(y_1,...,y_n)-\sum_{i=1}^nf(y_1,...,[x,y_i],...,y_n).$$ One can write $C^n$ as a quotient of $T^{n+1}(V)\otimes T^n(V^*)$ which is compatible with the action of $\mathfrak g$, here $V^*$ is the dual of $V$.
Question: How did he conclude that the eigenvalues of $\theta_x$ on $C^n$ are of the form $$(\lambda_1+...\lambda_{n+1})-(\gamma_1+...+\gamma_n) \text{ with }\lambda_i,\gamma_i\in L .$$
I think, in a rough sense, the action of $x$ on $C^n$ has eigenvalues $ \lambda_1...\lambda_{n+1}.\gamma_1^{-1}...\gamma_n^{-1}$, and so its derivative $\theta_x$ has eigenvalues of the above form. However, I could not write it explicitly, because I do not clearly understand the definition of $\theta_x$.
Here is the link to his paper: https://www.mathnet.ru/links/cbff0db3dfef43351a9052bd2de712ee/im2047.pdf
Aaaah, this was one of those things which I never had the pleasure to really write down in the detail it deserves. Thought this would be shorter, but to really lay it all out, you do require a bunch of subtle arguments along the way. Not all of it may be useful to you, but perhaps someone else will get something out of it!
Natural Actions
It may be helpful to clarify what people usually mean with the word natural in "natural Lie algebra action on $T^n(V)\otimes T^n(V^*)$". It is a general fact that if you have an action $\theta$ of a Lie algebra $\mathfrak g$ on a vector space $V$, then you automatically have:
Why do we call these actions natural? For one, they are canonical in the sense that we did not have to make any additional choices: this works for any Lie algebra representation without any additional input. This choice also makes some very natural maps into intertwiners, such as pairing a functional with an element of $V$: That gives us a map $$\langle \cdot \rangle : V \otimes V^* \to K, \quad v \otimes \alpha \mapsto \alpha(v).$$ We have $$\langle \theta_x^{12} (v \otimes \alpha) \rangle = \langle \theta_x v \otimes \alpha \rangle + \langle v \otimes \theta_x^* \alpha \rangle =\alpha(\theta_x v) + (\theta_x^* \alpha) (v) = 0.$$ But the field $K$ is usually considered to be equipped with the trivial Lie algebra action $\theta_x^\mathrm{trivial} k = 0$ for all $x \in \mathfrak g, k \in K$. This proves that $\theta_x^\mathrm{trivial} \langle v \otimes \alpha \rangle = \langle \theta_x^{12} (v \otimes \alpha) \rangle$, meaning the pairing intertwines the actions.
This combination of actions also reproduces adjoint actions: The natural action on the endomorphism algebra $\mathrm {End} (V) \cong V \otimes V^*$ becomes an endomorphism with, for all $w \in W$: $$\left\langle\left(\theta_x^{\mathrm{End}(V)} (v \otimes \alpha) \right), w\right\rangle = \left\langle\left((\theta_x v) \otimes \alpha\right), w\right\rangle + \left\langle\left(v \otimes ( \theta_x^* \alpha)\right), w\right\rangle = \theta_x(v \alpha(w)) - \alpha(\theta_x w) v = \langle \left(\theta_x \circ (v \otimes \alpha) - (v \otimes \alpha) \circ \theta_x)\right), w \rangle.$$
Hence we see that $\theta_x^{\mathrm{End}(V)} \phi = \theta_x \circ \phi - \phi \circ \theta_x$ reproduces commutator of endomorphisms. Exactly what we want the adjoint action to be!
Eigenvalues of Associated Actions
Let us check how eigenvalues of $\theta_x$ propagate to the dual and tensor product actions. For this, assume that $V, V_1, V_2$ are all finite-dimensional.
The dual action: $\lambda$ is an eigenvalue of $\theta^*_x$ iff $-\lambda$ is an eigenvalue of $\theta_x$. One can prove this, for example, by choosing a basis $\{e_i\}$ of $V$ and a corresponding dual basis $\{e^i\}$ of $V^*$: The matrix representation of $\theta_x$ in the basis will be exactly the negative transpose of the matrix representation of $\theta_x^*$ in the dual basis, so their Jordan normal forms will also only differ by a minus on the diagonal. Hence same eigenvalues up to a minus.
The tensor product action: All eigenvalues of $\theta^{12}_x$ are of the shape $\lambda_1 + \lambda_2$ where $\lambda_i$ are eigenvalues of $\theta_x^i$. Again, choose bases $\{e^1_i\}$ and $\{e^2_i\}$ of $V_1$ and $V_2$. If the representations of $\theta^i_x$ in the basis of $V_i$ are matrices $M_i$, then $\theta^{12}_x$ has the matrix representation $M_1 \otimes I_{\dim V_2} + I_{\dim V_1} \otimes M_2$ in the product basis $\{e_i^1 \otimes e_j^2\}$, where the $I$ are identity matrices. $M \otimes I$ has the same eigenvalues as $M$ which is easy to check. Also, $M_1 \otimes I$ commutes with $I \otimes M_2$, so they are simultaneously triangularizable (if the field is algebraically closed! But this restriction does not concern us, we can always go to the algebraic closure for our arguments about eigenvalues). Putting them into upper triangular form, the diagonals of both matrices simply add up, and since we can read the eigenvalues of a matrix off its upper triangular form, we have shown that the eigenvalues of $\theta^{12}_x$ are of the desired shape: $\lambda_1 + \lambda_2$ where $\lambda_i$ are eigenvalues of $\theta^i_x$.
Eigenvalues and Invariant Subspaces
Now, to bring it back to your question: The Lie derivative that you wrote above is natural in exactly this sense that it comes from the above natural actions. That is pretty good, because we know now how the eigenvalues of the original action perpetuate through the dualizations and tensor products. As an action on $T^n (V) \otimes T^m (V^*)$ we indeed get that $\theta_x$ has eigenvalues of the shape $$\lambda_1 + \dots + \lambda_n - \nu_1 - \dots - \nu_m$$ for eigenvalues $\lambda_i, \nu_j$ of $\theta_x$. The last thing to do is to see what happens when we lift this action to a quotient by a $\theta_x$-invariant subspace. Let's abstractly say we have a linear endomorphism $A : V \to V$ on a finite-dimensional vector space $V$, and there is a subspace $W \subset V$ to which $A$ restricts, so as an endomorphism $A|_W : W \to W$. By choosing a basis of $W$ and extending it to a basis of $V$, we can find a matrix representation of $A$ which is in block shape $$A = \left[ \begin{array}{c|c} A|_{V/W} & * \\ \hline 0 & A|_W \end{array} \right].$$ By the Leibniz formula, the characteristic polynomial $\chi_A$ of $A$ factorizes into $\chi_A = \chi_{A|_{V/W}} \chi_{A|_W}$. Hence the eigenvalues of the quotient map $A|_{V/W} : V/W \to V/W$ are a subset of the eigenvalues of $A$. In our specific case, that means that the eigenvalues of the Lie algebra action on the quotient are also of the shape $\lambda_1 + \dots + \lambda_n - \nu_1 - \dots - \nu_m$, which is what we wanted to show.
From Group Actions to Lie Algebra Actions
Your misunderstanding may have come from the fact that the formula for the tensor product of a Lie algebra action is not $x \cdot (a \otimes b) = (x \cdot a) \otimes (x \cdot b)$. If you are used to group actions instead, that would be the right formula. But on second thought, it all makes sense: Lie algebras are to be seen as the infinitesimal versions of Lie groups, and accordingly Lie algebra actions should act like derivatives of Lie group actions. Let's say we have a Lie group action $\theta^G$ and want to define a Lie algebra action as a derivative thereof, $\theta^{\frak g}_x := \frac{d}{dt}|_{t=0} \theta^G_{\exp(tx)}$. One uses the exponential map $\exp$ to go from $\frak g$ to a Lie group $G$, so by a product rule argument: $$ \theta^{\frak g}_x \cdot (v_1 \otimes v_2) = \frac{d}{dt}|_{t=0} \theta^G_{\exp(tx)} (v_1 \otimes v_2) = \frac{d}{dt}|_{t=0} ( \theta^G_{\exp(tx)} v_1) \otimes (\theta^G_{\exp(tx)} v_2) = \left( \frac{d}{dt}|_{t=0}\theta^G_{\exp(tx)}v_1\right)\otimes v_2 + v_1 \otimes \left(\frac{d}{dt}|_{t=0}\theta^G_{\exp(tx)} v_2 \right) = (\theta_x^{\frak g} v_1) \otimes v_2 + v_1 \otimes (\theta_x^{\frak g} v_2). $$
This, together with the tensor product action being compatible with pairings and endomorphisms as we discussed earlier really consolidates that this is the "right" way to act on a tensor product.