Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose V is a two-dimensional vector space.

According to Harris' "Algebraic Geometry", p.118 the vectors in $\mathbb{P}(\operatorname{Sym}^2(V))$ of the form $v\cdot v$ are supposed to be exactly the elements of $v_2(\mathbb{P}^1)$, which is the image of the quadratic Veronese map.

However, I fail to see this correspondence. If I take basis vectors $v_1, v_2$ of V and suppose that $v=\lambda_1v_1 + \lambda_2v_2$, wouldn't $v\cdot v$ just be equal to

$\lambda_1^2(v_1\cdot v_1) + 2\lambda_1\lambda_2(v_1\cdot v_2)+\lambda_2^2(v_2\cdot v_2)$ in $\operatorname{Sym}^2(V)$,

and thus correspond to $[\lambda_1^2:2\lambda_1\lambda_2:\lambda_2^2]$ in homogeneous coordinates (which is not in the image of $v_2$)?

I assume that I have either just miscalculated or severely misunderstood some central concept. Either way, I'd be grateful for some help.

share|cite|improve this question
I ran into this same issue some time ago. The only way I could think of to resolve it was to say that the image of the vectors $v \cdot v$ is isomorphic to the image of the Veronese map, but not equal to it. You have to compose the map you give with the map $[x:y:z] \mapsto [x:y/2:z]$ to get the image of the quadratic Veronese. The same is true for higher degree Veronese maps ($v \cdot v \cdot \ldots \cdot v$ in $\mathbb{P}(Sym^n(V))$ is isomorphic to $v_n(\mathbb{P}^1))$, but you have to follow the obvious map with a linear map involving multinomial coefficients. – Ted Oct 22 '11 at 18:27
That sounds reasonable. I guess I was expecting a "stronger" form of isomorphy between the two sets, but this should be enough to prove the whole point of this paragraph ($PGL_2(K)$ being the automorphism group the quadratic veronese). Thanks! – Cedric B Oct 22 '11 at 18:37
In characteristic zero at least, the Veronese map should be defined by the map $[v] \mapsto [v \cdot v]$ and so you should have all the multinomial coefficients running around. Since rescaling coordinates does not have any geometric effect, it is customary to lazily but harmlessly drop the multinomial coefficients when coordinatizing the natural map $[v] \mapsto [v \cdot v]$. – Michael Joyce Oct 22 '11 at 19:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.