# Why is the image of the Veronese map defined by equations $X_I X_J = X_K X_L$?

The Veronese map is defined as $$v_d : \mathbb{P}^n → \mathbb{P}^{{n+d \choose n }−1}$$ where $$[z_0, . . . , z_n]$$ maps to $$[(z_0^{ i_0}· · · z_n^{ i_n} )_{0≤i_k≤d,\sum_{k=0}^ni_k=d}].$$

The image of the map lies on the variety described by equations

$$X_I · X_J − X_K · X_L = 0$$

where $$I + J = K + L$$ as multi-indices. The difficult part is to prove the converse: if the equations are satisfied, then $$X$$ is in the image of $$v_d$$. The proof of this fact in algebraic geometry books I came across begins with the observation that $$X_{(0, \ldots, d, \ldots, 0)} \ne 0$$ for some multi-index of type $$(0, \ldots, d, \ldots, 0)$$ (e.g. see page 32 here). Then the map $$u$$ such that $$u \circ v_d = \mathrm{Id}_{\mathbb{P_n}}$$ is easily constructed for the chart $$X_{(0, \ldots, d, \ldots, 0)} \ne 0$$.

But even if we prove that the map is well defined on the intersection of charts, I fail to see why it is the actual inverse map, i.e. why is $$v_d \circ u = \mathrm{Id}_{\mathbb{P}^{{n+d \choose n }−1}}$$

Given what is proved, that is, of course, equivalent, to showing that the image of $$v_d$$ is defined by equations $$X_I · X_J − X_K · X_L = 0$$. But that is the difficult part we were trying to prove in the first place, which just seems to be avoided in all algebraic geometry texts I came across. Also note that similar questions on stackexchange have the same problem. I have given my "elementary" proof, but I would be grateful if someone can explain how it follows from the proof outlined above, e.g. by some general theorem about homomorphisms.

## 2 Answers

Indeed, there are truly two things that need to be shown: that the Veronese map is an embedding, in particular injective (i.e. a one-sided inverse function exists) and that the image is defined by the given quadratic equations.

The injectivity comes from the fact that if you set one of the projective coordinates on the input point to be $$1$$, then its coordinates literally occur as a subset of the coordinates of the image point. (The remaining coordinates are monomials in those first few coordinates.) This is more or less the argument you described above.

This is also the source of the quadratic equations indexed by $$I+J=K+L$$: if you specialize $$I = (d,0,\ldots,0)$$ on the chart where the corresponding coordinate is $$1$$, the equation simply asserts that $$X_J$$ is the product of two other coordinates.

To show that these equations define the image, the argument is: given a point satisfying the equations, we pick a chart as you said. Next we use the first few coordinates of $$\mathbb{P}^N$$ ($$N$$ is that binomial coefficient) to write down the appropriate point of $$\mathbb{P}^n$$. (This implicitly applies the injectivity argument, which as you noted is easier). Finally, the quadratic relations state that the remaining projective coordinates are in fact the appropriate monomials in the first few coordinates, i.e. the point of $$\mathbb{P}^N$$ is indeed the image of the constructed point.

Aside: The same overall argument also proves the equations for Segre embeddings (just change “monomial” to “product”) and Plücker embeddings of Grassmannians (change “monomial” to “minor” or simply “polynomial”; here the input coordinates should be thought of as entries of a matrix with a certain block normalized to an identity matrix).

• I think this is a great explanation for what is really going on in the inductive arguments. Dec 11, 2022 at 8:51

I can prove in the straightforward manner, that once $$X_{(d, 0, \ldots, 0)}, X_{(d-1, 1, \ldots, 0)}, \ldots X_{(d-1, 0, \ldots, 1)}$$ have been fixed, through equations, they determine all other $$X_I$$ uniquely. Hence it follows that the map $$u$$ is indeed an inverse, because by construction $$v_d \circ u$$ will not change the components $$X_{(d, 0, \ldots, 0)}, X_{(d-1, 1, \ldots, 0)}, \ldots X_{(d-1, 0, \ldots, 1)}$$.

Proof Let $$I = (l, \ldots), l < d-1$$. Let's prove "by induction in reverse order" that $$X_I$$ is uniquely determined. We can clearly write $$X_I X_{(d, 0, \ldots, 0)} = X_{(l+ 1, \ldots)} X_{(d - 1, 0, \ldots, 1, \ldots, 0)}$$, so dividing the RHS by $$X_{(d, 0, \ldots, 0)}$$ we determine $$X_I$$.