# What is the total polarization of the determinant?

Let $A\in\mathfrak{gl}(\mathbb{R},n)$ be an endomorphism, and think up to conformal factors (in particular, $\Lambda^n\mathbb{R}^n$ will be the same as $\mathbb{R}$). By the total polarization $\widehat{p}$ of a homogeneous polynomial $p\in S^k V$, where $V$ is a vector space, I mean the image of $p$ via the embedding $S^k V\subset V^{\otimes k}$. E.g., if $Q$ is a quadratic form, $B:=\widehat{Q}$ is the associated bilinear form.

"THEOREM" A. There exists a unique unitary graded algebra endomorphism $\varphi_A$ of $\Lambda^\bullet\mathbb{R}^n$ such that $\left.\varphi_A\right|_{\mathbb{R}^n}=A$.

"THEOREM" B. There exists a unique graded derivation $\delta_A$ of degree 0 of $\Lambda^\bullet\mathbb{R}^n$ such that $\left.\delta_A\right|_{\mathbb{R}^n}=A$.

As a corollary, we get $\det\, A= \left.\varphi_A\right|_{\Lambda^n\mathbb{R}^n}$ and $\mathrm{tr}\, A=\left.\delta_A\right|_{\Lambda^n\mathbb{R}^n}$.

QUESTION: what is known about $\widehat{\mathrm{det}}$? Should be a very fundamental object, but I never heard about it. Moreover, it is possible to characterize it intrinsically, in analogy with "THEOREM" A above?

Let me add a few comments. In the case $n=2$ the answer is known (I've just learned it from a very beautiful MO answer---look here: https://mathoverflow.net/a/32583/22606): $\widehat{\mathrm{det}}$ is the Killing form. But what happens as $n$ grows larger? In fact, my question is just the tip of the iceberg, since I may extend it to all the coefficients $\psi_k(A)$ of the characteristic polynomial $p_A(t)$ of $A$, understood as homogeneous polynomials on $\mathfrak{gl}(\mathbb{R},n)$: how to describe their total polarizations?

Of course, I'm able to produce a brutal answer, namely $$\widehat{\psi_k}:(A_1,\ldots,A_k)\in \mathfrak{gl}(\mathbb{R},n)^{\otimes k}\longmapsto \mathrm{tr}\,(A_1\wedge\cdots\wedge A_k)\in\mathbb{R},$$ but then I'm not able to justify the symmetry of $\widehat{\psi_k}$, and the very definition of $A_1\wedge\cdots\wedge A_k$, which should be an endomorphism of $\Lambda^k\mathbb{R}^n$, is unclear. My guess is that the last formula can be made "clean" by combining the above defined $\varphi_{A_i}$'s and $\delta_{A_i}$'s, but guessing is not enough, and it's all I can do!

EDIT: After reading prof. Michor's answer, I understood that $\widehat{\psi_k}$ can be though of as the trace of the symbol of the $k^\textrm{th}$ order differential operator $\delta_{A_1}\circ\cdots\circ\delta_{A_k}$, restricted to the $k^\textrm{th}$ homogeneous component $\Lambda^k\mathbb{R}^n$ (this point of view was suggested by L. Vitagliano). This way, one achieves total symmetry without averaging on the permutation group, but solely relying on the properties of the symbol instead. However, until such time a practical use of this - in many respects - cumbersome point of view is found (like, e.g., a new proof go Hamilton-Cayley theorem), I guess the answer below if the most straightforward one.

Your brutal answer is slightly wrong. Namely, your map $(\mathbb R^n)^{\otimes k}\to \Lambda^k(\mathbb R^n)$ which is given by $$x_1\otimes\dots\otimes x_k \mapsto A_1x_1\otimes\dots\otimes A_kx_k\mapsto A_1x_1\wedge\dots\wedge A_kx_k$$ does not factor over $x_1\otimes\dots\otimes x_k\mapsto x_1\wedge\dots\wedge x_k$. But $$x_1\otimes\dots\otimes x_k\mapsto \frac1{k!}\sum_{\sigma\in S_n}\text{sign}(\sigma) A_1x_{\sigma1}\wedge\dots\wedge A_kx_{\sigma k} =\frac1{k!}\sum_{\sigma\in S_n} A_{\sigma1}x_1\wedge\dots\wedge A_{\sigma k}x_k$$ does, and the trace of this is the polarization of the $k$-th characteristic coefficient.