I'm following a course in algebraic geometry and I'm asked to write a short expository paper about the topic that every nonsingular curve can be embedded in $\mathbb{P}^3$. You should be aware that I have almost no background in algebraic geometry. When I assume that the curve $X$ lies already in some $\mathbb{P}^m$ it is doable. It is done by taking a point $P \in \mathbb{P}^m \backslash X$ outside the curve and project the curve from the point onto a hyperplane $H \cong \mathbb{P}^{m-1}$, this works when $m \geq 4$. The conditions $P$ has to fulfil are the following:
- $P$ is not on any secant line of $X$,
- $P$ is not on any tangent line of $X$.
By my very limited background, it would help a lot for me to understand the algebraic geometry involved if someone could give an elaborated example of a curve lying in $\mathbb{P}^4$ and project it in $\mathbb{P}^3$.
Edited. (Many thanks to Mohan for his comment!)
Let's work over the field of complex numbers, and let's consider the curve in $\mathbb{CP}^4$ defined by the parametrisation: $$ t \in \mathbb C \cup \{ \infty \} \ \ \ \mapsto \ \ \ [x_0 : x_1: x_2: x_3: x_4] = [t^4 : t^3 : t^2 : t : 1 ] \in \mathbb{ CP}^4.$$
(This curve is of course isomorphic to $\mathbb {CP}^1$: it is the image of $\mathbb{CP}^1$ under the fourth Veronese map.)
Example 1: We can try projecting from the point $[0:0:1:0:0]$ onto the hyperplane $x_2 = 0$. The image of our curve under this projection can be parametrised as follows: $$ t \in \mathbb C \cup \{ \infty \} \ \ \ \mapsto \ \ \ [x_0: x_1: x_3 : x_4] = [t^4 : t^3 : t : 1 ] \in \mathbb{ CP}^3.$$ This projection is isomorphic to the original curve. By performing this projection, we have succeeded in embedding our curve in $\mathbb{CP}^3$.
Example 2: We start with the same curve, but this time, we will try projecting from the point $[0:0:0:1:0]$ onto the hyperplane $x_3 = 0$. The image under this projection can be parametrised as: $$ t \in \mathbb C \cup \{ \infty \} \ \ \ \mapsto \ \ \ [x_0 : x_1: x_2: x_4] = [t^4 : t^3 : t^2 : 1 ] \in \mathbb{ CP}^3.$$ But this is not isomorphic to our original curve! It has a cusp at $t = 0$, whereas our original curve was smooth. The reason why a cusp appears under this projection is that our point of projection $[0:0:0:1:0]$ lies on the tangent to the curve at $t = 0$. Therefore, if you imagine that you're walking along the curve, then at $t = 0$, your "shadow" under the projection is moving at infinitesimally small velocity, hence the appearance of the cusp.
Example 3: This time, we will try projecting from the point $[1:1:1:1:0]$ onto the hyperplane $x_0 + x_1 + x_2 + x_3 = 0$. First, to make our lives easier, we'll define a new basis for the $\mathbb C^5$: $$ y_0 = x_0 + x_1 + x_2 + x_3, \ \ \ \ y_1 = x_0 - x_3, \ \ \ \ y_2 = x_1 - x_3, \ \ \ \ y_3 = x_2 - x_3, \ \ y_4 = x_4.$$ Using this new basis, the image of our curve under the projection is parametrised as: $$ t \in \mathbb C \cup \{ \infty \} \ \ \ \mapsto \ \ \ [ y_1: y_2: y_3: y_4] = [t^4 - t: t^3 - t : t^2 - t : 1] \in \mathbb{ CP}^3.$$ Again, we have a problem: $t = 0$ and $t = 1$ map onto the same point under the projection, so the image of our curve under the projection has a node, whereas the original curve did not. And why does this node appear? Because the secant joining the points $t = 0$ and $t = 1$ on the original curve passes through the point of projection $[1:1:1:1:0]$; in other words, $t = 0$ and $t = 1$ cast the same "shadow" under the projection.