I'm currently following an introductory course on algebraic geometry. When studying projective varieties, which we defined as $V(f)=\{ [x] | f[x]=0 \}$ given that f is a homogeneous polynomial, the following definition of tangent cones came up: If $\mathbb{C}P^n$ is a projective space and the projective variety $V(f) \subset \mathbb{C}P^n$, then we define the tangent cone with vertex $p \in V(f)$ as $C=V(\sum \frac{\delta^k f}{\delta x_{i_1}...\delta x_{i_k}}(p) x_{i_1}...x_{i_k})$ for every $x_{i_1}...x_{i_k}$ and k is the smallest integer for which there exists an $x_{i_1}...x_{i_k}$ so $ \frac{\delta^k f}{\delta x_{i_1}...\delta x_{i_k}}(p) \neq 0$. My question is this: Why is C a cone, ie. why will $pq \subset C$ if $q \in C$?
Thanks!
Let's look at this on an affine patch; so assume that we're looking at $\{f=0\}$ in $\mathbb{A}^n$. Let's assume that the point we're looking at is $(0,0)$; if not, we do a linear change of coordinates. It is easy to see then that your definition of tangent cone is equivalent to looking at the zero set of the homogeneous part of $f$ of least degree. So, for example, if $f=y^2-x^2-x^3$, then the homogeneous part of least degree is $y^2-x^2$, and so the tangent cone is $y^2-x^2=0$; that is, $(y-x)(y+x)=0$.
Since it is defined by a homogeneous polynomial (let's say $G$), then if $x$ is in the tangent cone, we have that $G(\lambda x)=\lambda^dG(x)=0$, where $\lambda\in \mathbb{C}$.
This shows that it is a cone in the sense you're saying; now be careful, it is not a convex cone!