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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.

  • 1
    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 $[v] \mapsto [v \cdot v]$.2011-10-22

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