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As a preface I don't much background in this area and I think I am dealing with older (or currently non-standard) definitions.

Definition: A polynomial $f \in \mathbb{R}[x_1, \ldots, x_n]$ is homogenous of degree $d$ if for all $\lambda \in \mathbb{R}, f(\lambda x_1, \ldots, \lambda x_n) = \lambda^d f(x_1, \ldots, x_n)$.

Definition: An ideal $I \subset \mathbb{R}[x_1, \ldots, x_n]$ is homogenous if it is generated by finitely many homogenous polynomials.

Definition: $X \subset \mathbb{P}(\mathbb{R}^n)$ is a real projective variety if it is the zero locus of a prime homogenous ideal $I \subset \mathbb{R}[x_1, \ldots, x_n]$.


Now for my question, I have a $X \subset \mathbb{P}(\mathbb{R}^n)$ which is a real projective variety. I have a polynomial map $F = (f_1, \ldots, f_k)$ where $f_i : \mathbb{R}^n \to \mathbb{R}$ are homogenous monomials of degree $d$, with $d < n$. I am interested in understanding the image $F(X)$. In particular, I care about when $F(X)$ is closed with respect to the Zariski Topology on $\mathbb{P}(\mathbb{R}^k)$.

From reading I know that if we were working over an algebraically closed field then $F$ is a closed and hence $F(X)$ is a variety, as discussed here. However, in my case we are working over $\mathbb{R}$. I cannot seem to find anything about the case when the field is not algebraically closed.


For more context, for a given $m$, I have matrices $(A_1, \ldots, A_m) \in \mathbb{P}(\mathbb{R}^{m \times M \times N})$ such that $A_i$ have rank at most $r$ (can assume $r=2$). Hence $(A_1, \ldots, A_m)$ live in a projective variety. I interested in the map

$$ (A_1, \ldots, A_m) \mapsto A_1 \odot \ldots \odot A_m$$

where $\odot$ is the Hadamard product.

1 Answers1

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In general the image of a "morphism" in real algebraic case is semi-algebraic. However, in your case the map is (probably)finite, so proper, hence it has a closed ( in usual topology) image. So it is (probably) defined as a finite union of sets of the form $\{F_1\ge 0, \ldots F_k\ge 0\}$. Explicitely it might get very involved.

orangeskid
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  • Past identifying that semi-algebraic geometry deals with this sort of problem, I don't think this is a great answer: in the usual topology, any continuous map from a compact space to a Hausdorff space is topologically proper, so it will have image closed in the usual topology. Further, because the map is semi-algebraic, the image is already a semi-algebraic set, so something like the form you write will be automatic (there is potentially some care to be taken here about viewing $\Bbb RP^n$ as a subset of $\Bbb R^N$ and picking polynomials there in order versus coordinates on $\Bbb RP^n$...). – KReiser Jun 29 '22 at 04:29
  • @KReiser: Thank you for your observations! I mentioned finite and proper because sometime the map is defined on affine varieties. Now, for projective maps I am pretty sure that the image will be a union of elementaries as described above, may be standard, but I don't have a source. Where inequalities occur they are also homogeneous of even degree. So things should be OK on the projective target. – orangeskid Jun 29 '22 at 04:42
  • Thanks for the response. Any references to how I could compute these elementaries or determine any properties for these elementaries for my concrete map would be appreciated. – Rishi Sonthalia Jun 29 '22 at 17:26
  • In particular, (this is not in my original question) but I am interested in how these elementaries change or don't change as I vary $m$. – Rishi Sonthalia Jun 29 '22 at 17:34
  • @Rishi Sonthalia: No worries! I have a question about the definition: in Hadamard product you can get the product $0$, while the matrices have full rank. But then to define it from a projective space, you need it to be defined, so perhaps to land in a projective space. All in all, I think the source cannot be considered projective. – orangeskid Jun 29 '22 at 17:49
  • I am not sure I fully understand, but in my setting the matrices actually cannot have $0$s. This actually comes from a data problem that I am trying to model with some math! – Rishi Sonthalia Jun 29 '22 at 18:14
  • @Rishi Sonthalia: I see, no zeroes makes it a subset of projective space, including the image and all can be done projectively. Certainly the case $r=1$ is simple, you get again a matrix of rank $1$. Hadamard multiplication with a matrix of rank $1$ does not change the rank. So $r=2$ is the next case, two $3\times 3$ matrices of rank $2$, what can you say about their Hadamard product? Interesting. – orangeskid Jun 29 '22 at 18:22