Correspondence between bitangents of a quartic and odd theta characteristics

Question

Let $C$ be a Riemann surface of genus $g=3$. I can't understand why the following statements are true:

1. If $C$ is not hyperelliptic, then the canonical series $|K|$ embeds $C$ as a smooth quartic in $\mathbb{P}^2$
2. The odd theta characteristics of $C$ correspond to the bitangent lines to this quartic.

How can these facts be explained?

Answer 1

Since for a canonical divisor $K$ the linear system $|K|$ has no base point as soon as $g\geq1$ ( this results from Riemann-Roch), the canonical map $i_K:C\to \mathbb P^2$ is an everywhere defined morphism.
The only way it could fail to be an embedding is if for some $p,q\in C$ (not necessarily distinct) we had $\dim_\mathbb C \Gamma(C,\mathcal O(p+q))=2$ : this follows from Riemann-Roch too.
But this is equivalent to $C$ being hyperelliptic.
In the other cases $i_K$ is an embedding and the degree of its (smooth) image is the degree of the divisor $K$, namely $2g-2=2.3-2=4$.
You can find supplementary details on page 247 of Griffiths-Harris's Principles of Algebraic Geometry.