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

Let $V$ be a vector space over $\mathbb{C}$ and $V^*$ be its dual space. Let $\mathbb{P}V$ be the projective associated to $V$. It is said that homogeneous coordinates of $\mathbb{P}V$ correspond to elements of $V^*$. What is the correspondence? Suppose that $V=\langle v_1, \ldots, v_k \rangle$ and $V^*=\langle v^*_1, \ldots, v^*_k \rangle$ where $v^*_i(v_j)=\delta_{ij}$. We can let $(a_0, \ldots, a_k) \in \mathbb{P}V$ corresponds to $f\in V^*$ where $f(v_i)=a_i$. Is this the correct correspondence? Thank you very much.

share|cite|improve this question
The duality between $V$ and $V^*$ induces a "projective duality" between $\mathbb{P}V$ and $\mathbb{P}(V^*)$. For example, projective hyperlanes in $\mathbb{P}V$ correspond to vector hyperplanes in $V$. Such an hyperplane is the kernel of non zero linear form $f\in V^*$ unique up to a constant. In other words it corresponds to a line in $V^*$, i.e. a point in $\mathbb{P}(V^*)$. – AFK Sep 7 '11 at 17:04
up vote 1 down vote accepted

I think I misunderstood your question in my comment. Let me try and correct that.

Let $V$ be a finite dimensional $k$-vector space. A set of linear coordinates on $V$ is simply a basis $(\alpha_i)$ of $V^*$. Taking the coordinates of a vector $v\in V$ is considering its image under the isomorphism $\alpha: V \to k^n$, $v \mapsto (\alpha_1(v),\ldots,\alpha_n(v))$.

Now $\mathbb{P}(V) = (V\setminus \{0\})/k^\times$ and homogenous coordinates on $\mathbb{P}(V)$ are just linear coordinates on $V$ up to the action of $k^\times$. The map $\alpha: V \to k^n$ above induces $\bar{\alpha} : \mathbb{P}(V) \to (k^n\setminus \{0\})/k^\times = \mathbb{P}^n$. So a set of homogenous coordinates corresponds to a basis $(\alpha_i)$ of $V^*$ (and not just one element $f\in V^*$ as you seem to think).

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.