Vtyjkyuk

Let $E$ a $mathbb R-$vector spaces of dimension $n$. We say that $f$ is $p-$alternating if $$f(x_1,...,x_p)=0,quad (x_1,...,x_p)in E^p$$
whenever $x_i=x_j$ for some $ineq j$ (and $pleq n$). Since $mathbb R$ has a characteristic $neq 2$ this is equivalent to $$f(x_sigma 1,...,x_sigma p)=varepsilon(sigma )f(x_1,...,x_p)$$
for all $sigma in mathfrak S_p$. And such linear form are called "antisymmetric". I know that with matrices, the set of matrices are in direct sum with symmetric and antisymmetric matrices, i.e. $$mathcal M_ntimes n(mathbb R)=mathcal S_noplus mathcal A_n.$$

Question : If I denote $mathcal L_p(E)$ the space of linear form $Eto mathbb R$, $Lambda _p(E)$ the space of $p-$linear form and $mathcal S_p(E)$ the space of symmetric linear form, does it hold that $$mathcal L_p(E)=Lambda _p(E)oplus mathcal S_p(E) ?$$

Attempts

The thing is I don't find any theorem that says that. If $fin mathcal L_p(E)$, then if I denote $$f_ij(x_1,...,x_i,...,x_j,...,x_p)=f(x_1,...,x_j,...,x_i,...,x_p)$$
$$f(x_1,...,x_p)=underbracefracf+f_ij2_:=g+underbracefracf-f_ij2_:=htag*$$
then $g$ is symmetric and $h$ antisymmetric. So
$$mathcal L_p(E)=Lambda _p(E)+mathcal S_p(E) $$
looks true. But the the writing $(*)$ looks far to be unic, so I have big doubt that the sum is direct. Also, I tried : let $fin mathcal S_p(E)cap Lambda _p(E) $.
Then $f(x_1,...,x_p)=f_ij=-f_ij$ anf thus $$2f_ij=0implies f_ij=0,$$
therefore $f=0$. So it looks to be a direct sum. Since I've never sen such a result, I have doubt that my argument is correct.

$newcommandRmathbb R$Your $f+f_ij$ and $f-f_ij$ are symmetric and anti-symmetric only with respect to the $i$-th and $j$-th argument. They are not symmetric and anti-symmetric as multilinear forms.

Note that the dimensions of $mathcal L_p(R^n)$, $Lambda_p(R^n)$ and $mathcal S_p(R^n)$ are $n^p$, $binomnp$ and $binomn+p-1p$, respectively. What does this say about your conjecture that $mathcal L_p(R^n)=Lambda_p(R^n)oplusmathcal S_p(R^n)$?