5
$\begingroup$

Let $M_{2,4}(\mathbb{R})$ be the set of real $2\times4$-matrices of rank $2$. For any $A\in M_{2,4}(\mathbb{R})$ and $1\leq i<j\leq 4$, let $p_{ij}$ be the corresponding $2\times 2$-minors of $A$. The image $K_{2,4}$ of the Plücker map $$ \mathcal{P}: M_{2,4}(\mathbb{R})\to\mathbb{R}^6,\;\;\; \mathcal{P}(A)=(p_{12}, p_{13}, p_{14}, p_{23},p_{24},p_{34}) $$ is the affine cone over the Klein quadric $Gr_2(\mathbb{R}^4)\hookrightarrow\mathbb{P}^5(\mathbb{R})$ given by the famous Plücker relation $p_{12}p_{34} - p_{13}p_{24} + p_{14}p_{23} = 0$.

Now, I am interested in finding "nice" (continuous/smooth/analytic...) functions $f:\mathbb{R}\to\mathbb{R}$, such that the map $F(x_1,\ldots,x_6) := (f(x_1),\ldots,f(x_6))$ preserves $K_{2,4}$.

I have the following partial example: let $K^0_{2,4} = \mathcal{P}(M^0_{2,4}(\mathbb{R}))$, where $M^0_{2,4}(\mathbb{R})\subset M_{2,4}(\mathbb{R})$ is the set of matrices that annihilate the column vector $(1,1,1,1)^t$. Then for $f(x)=A\sin(\alpha x)$ and $f(x)=A\sinh(\alpha x)$ we have $F(K^0_{2,4})\subset K_{2,4}$. It would be also nice to have other examples of functions with this last property.

$\endgroup$
0

1 Answer 1

4
$\begingroup$

Restricting to $M^0_{2,4}(\mathbb{R})$ implies that $$0 = p_{12}+p_{13}+p_{14} = p_{12}-p_{23}-p_{24} = p_{13}+p_{23}-p_{34}$$ is the $p_{ij}$ linear relations. The quadratic relation is then $$0=p_{12}(p_{13}+p_{23})-p_{13}(p_{12}-p_{23})+(-p_{12}-p_{13})p_{23}$$ which this an instance of the algebraic identity $0=a(b+c)-b(a-c)-c(a+b).$ The analytic solutions to the functional equation $0=f(a)f(b+c)-f(b)f(a-c)-f(c)f(a+b)$ are just $f(x)=ax$ and $f(x)=a\sin(bx)$, for any complex constants $a,b$. This can be proved by looking at power series coefficients of the functional equation given the power series of $f(x)$.

$\endgroup$
1
  • $\begingroup$ Thank you. This is very expected in the analytic category. I suppose you just use the power series substitution to deduce the result. $\endgroup$
    – Serj
    Commented Sep 7, 2017 at 7:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .