Image of the Veronese is defined by $X_I X_J=X_k X_L=0$

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I am going through the notes https://www2.math.ethz.ch/education/bachelor/lectures/fs2014/math/alg_geom/AG_lecture_notes.pdf

He defines the Veronese embedding of degree $d \in N$ as the map $ν_d : \mathbb{P}^n → \mathbb{P}^{{n+d \choose n }−1}$ 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}].$

He describes the following equations

Let $X_{i_0,...,x_n}$ be the variable in $\mathbb{P}^{{n+d \choose n }−1}$ that corresponds to the monomial $z_0^{ i_0} · · · z_n^{ i_n}$ in the Veronese map. Then, we have on the image of the Veronese map, for the multi-indices $I = i_0, . . . , i_n, J = j_0, . . . , j_n, K = k_0, . . . , k_n, L = l_0, . . . , l_n$ such that $I + J = L + K:$

$X_I · X_J − X_K · X_L = 0$

On page 11,

Proposition 5.4 claims the image of the Veronese is defined by the quadratic equations as described above.

I have trouble see this a proof for this. Any help is appreciated.

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Here is a sketch, let me know if you want more detail.

It is clear that the image of the Veronese satisfies the given equations. Conversely, you first want to show the lemma that more generally, $X_{I_1} \cdots X_{I_k} = X_{J_1}\cdots X_{J_k}$ provided $\sum_i I_i = \sum_i J_i$ is implied by the quadratic relations. This can be done by induction on $k$.

Then, let $X_I = a_I$ satisfy the quadratic equations. We must show it is the image of a point of $\mathbb{P}^n$. There is some $I$ so that $a_I \ne 0$. WLOG suppose $I_1 > 0$. Choose $k$ so that $k \ge \lceil d/I_1\rceil$. Then by our lemma $a_I^k = a_{(d,0,\ldots,0)} a_{I_2}\cdots a_{I_k}$, for some choice of $I_2,\ldots,I_k$. This implies $a_{(d,0,\ldots,0)} \ne 0$. Then the point $(1, \frac{a_{(d-1,1,0,\ldots,0)}}{a_{(d,0,\ldots,0)}}, \ldots, \frac{a_{(d-1,0,\ldots,0,1)}}{a_{(d,0,\ldots,0)}}) \in \mathbb{P}^n$ maps to the $a_I$, which can be seen by applying the lemma again.