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Sorry if this question is asked already, a quick google search didn't yield any answers.

In Classical algebraic geometry, §1.1.6, Dolgachev defines the Steinerian hypersurface $\operatorname{St}(X)$ of a given hypersurface $X \subset \mathbb{P}^n$ of degree $d$ as the locus of all singular points of polar quadrics of $X$. There is also the Hessian hypersurface $\operatorname{He}(X) = \bigcup_{a \in \operatorname{St}(X)}\operatorname{Sing}(P_a(X))$.

He writes:

Assume that the quadric $P_{a^{d−2}}(X)$ is of corank $1$. Then it has a unique singular point $b$ with the coordinates $[b_0, \dotsc, b_n]$ proportional to any column or a row of the adjugate matrix $\operatorname{adj}(\operatorname{He}(f))$ evaluated at the point $a$. Thus, $\operatorname{St}(X)$ coincides with the image of the Hessian hypersurface under the rational map

$$\operatorname{st}: \operatorname{He}(X) \dashrightarrow \operatorname{St}(X),\quad a \mapsto \operatorname{Sing}(P_{a^{d-2}}(X)).$$

Also, if the first polar hypersurface $P_a(X)$ has an isolated singular point for a general point $a$, we get a rational map

$$\operatorname{st}^{-1} : \operatorname{St}(X) \dashrightarrow \operatorname{He}(X),\quad a \mapsto \operatorname{Sing}(P_a(X)).$$

Dolgachev says it is a difficult problem to understand the loci of indeterminacy of this map (and the inverse, when it exists), so I had an idle curiosity to know what is known about this situation, depending on the dimension and other geometric features of $X$. Since a quick search on the arxiv didn't yield any immediate result, I thought I'd ask here, since maybe I don't have the relevant keywords, or maybe the answer is just 'nothing!'

In Dolgachev, it is assumed that the base field is $\mathbb{C}$, or at least characteristic $0$ and algebraically closed, but if changing those assumptions makes for something interesting, that's fine by me too.

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  • $\begingroup$ Your quote seemed a little garbled at the end in comparison to the version of Dolgachev's notes I found online (for example, the source and target of $\operatorname{st}$ were switched), so I edited to match what appears in those notes. $\endgroup$
    – LSpice
    Commented Aug 3 at 23:00

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