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Let $F$ be a homogeneous from in $\mathbb{R}[x_0, .., x_n]$. Then $F$ defines a projective variety $X \subset \mathbb{P}_{\mathbb{C}}^n$. Assume $X$ is smooth. In this case $F=0$ also defines a submanifold $M = \{ \mathbf{x} \in \mathbb{R}^{n+1} \backslash \{ \mathbf{0} \} : F(x_0, .., x_{n+1}) = 0 \}$ of $\mathbb{R}^{n+1}$.

I was wondering if there were conditions I can impose on $F$ or $X$ such that $M$ has nowhere vanishing second fundamental form? Any comments are appreciated!

Edit: by nowhere vanishing second fundamental form I mean given any $p \in M$ the second fundamental form at $p$ does not vanish identically on $T_p M$.

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    $\begingroup$ Your $M$ is a cone, so contains lines (through the origin), and these are geodesics in the ambient space, so the second fundamental form vanishes on them. So you might clarify, do you mean that the second fundamental form is nonzero on some tangent vector, or on every nonzero tangent vector? A smooth quadric hypersurface clearly has homogeneous cone, so no tangent space has zero second fundamental form. $\endgroup$ – Ben McKay Nov 19 at 12:28
  • $\begingroup$ I have added an edit. Please let me know if the question still contains ambiguity. Thank you! $\endgroup$ – Takeshi Gouda Nov 19 at 14:00
  • $\begingroup$ Thanks, that's clear. $\endgroup$ – Ben McKay Nov 19 at 16:07
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    $\begingroup$ yes, the smooth quadric hypersurface is invariant under a group of symmetries which acts transitively on the real points of its cone, i.e. any two nonzero null vectors of a real quadratic form are carried to one another by a linear transformation preserving the quadratic form. The second fundamental form is a projective invariant, so its null space is too. The second fundamental form can't vanish everywhere, because the quadric is not linear. $\endgroup$ – Ben McKay Nov 19 at 17:41
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    $\begingroup$ The argument for the quadric is very special: there are very few homogeneous projective hypersurfaces. $\endgroup$ – Ben McKay Nov 20 at 10:18
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As we know, the projective hypersurface in $\mathbb{P}^n$ defined by a homogeneous polynomial equation $$ F(x^0,\ldots,x^n)=0 $$ of degree $m$ is nonsingular if $x=0$ is the only solution to the equations $$ 0 = F = \partial_0F = \partial_1F = \cdots = \partial_nF. $$ Because $mF = x^0\,\partial_0F + \cdots + x^n\,\partial_nF$, this is at most $n{+}1$ independent equations for $n{+}1$ unknowns. Generically, they are independent; hence, nonsingularity is a generic condition for a nonzero homogeneous polynomial $F$ (of any degree).

If one assumes that the hypersurface is nonsingular, then the points where the second fundamental form of the hypersurface vanishes are the places where $F=0$ and where the Hessian matrix $\mathrm{Hess}(F) = (\partial_i\partial_jF)$ belongs to the $(n{+}1)$-dimensional space of symmetric matrices of the form $$ \begin{pmatrix}\partial_0F\\\vdots\\ \partial_nF\end{pmatrix} \begin{pmatrix}v_0 & \cdots & v_n\end{pmatrix} +\begin{pmatrix}v_0\\\vdots\\ v_n\end{pmatrix} \begin{pmatrix}\partial_0F & \cdots &\partial_nF\end{pmatrix} $$ for general $v = \begin{pmatrix}v_0 & \cdots & v_n\end{pmatrix}$. These conditions are expressed by the satisfaction of ${n\choose 2}$ linear equations on $\mathrm{Hess}(F)$ (with coefficients that are quadratic in the $\partial_iF$), plus the equation $F=0$, of course. Thus, the places where the second fundamental form vanishes identically are the solutions of an overdetermined system of equations when $n>2$.

When $n=2$, this is only $2$ homogeneous equations for $3$ unknowns, so one expects that the generic smooth projective curve in $\mathbb{P}^2$ will have a finite number of flexes. Indeed, when the ground field is $\mathbb{C}$, then it is well-known that the only nonsingular curves without flexes are (nonsingular) conics.

When $n>2$, this is at least as many equations as unknowns, and, again, for the generic hypersurface, the locus of points where the second fundamental form vanishes identically will be empty.

It turns out that, if one asks for the condition that defines the points where the second fundamental form is merely degenerate (which, when $n>2$, is a weaker condition than vanishing identically), then it turns out that this locus is, generically, of dimension $n{-}2$ and, it seems, generalizing the case of $n=2$, that the only case of a smooth hypersurface when this locus is empty is a nonsingular hypersurface of degree $2$.

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  • $\begingroup$ Thank you very much for this very nice complete answer!! I was wondering 1) if you could possibly explain to me a little more or give reference on how I can show that the points where the second fundamental form vanishes is of the shape (the Hessian matrix belongs to the (n+1) dimensional space of symmetric matrices of the form...) as described above? $\endgroup$ – Takeshi Gouda Nov 22 at 18:44
  • $\begingroup$ 2) do you think it could possibly be the case that if $n$ is sufficiently large with respect to $\deg F$, then the locus of points where the second fundamental form vanishes identically will be empty, for any $F$ as in this question, not generically? $\endgroup$ – Takeshi Gouda Nov 22 at 18:48
  • $\begingroup$ Thank you for the clarification! I was thinking that if you could provide a sketch of what to do say for $n = 2$ or $3$, then I should be able to figure it out. $\endgroup$ – Takeshi Gouda Nov 23 at 11:59
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    $\begingroup$ @TakeshiGouda: I'm sorry, I see that there is a mistake in my formula for the cubic $F$. It should be $$F(x^0,x^1,\ldots,x^n) = (x^1)^3 + 3\,x^1\,(x^0)^2 + (x^2)^3 + \cdots + (x^n)^3.$$ (My typing was a bit hasty yesterday.) This $F$ has degree $3$ and is smooth. The second fundamental form of $F=0$ vanishes identically at $p=(1,0,\ldots,0)$. Similar examples can be given for any degree above $2$. For your first question, I can put in the argument, but it may take me a while. $\endgroup$ – Robert Bryant Nov 23 at 12:18
  • $\begingroup$ Just to make sure I understood everthing correctly could you please explain what you mean by 'generic condition' in this context? Thank you very much! $\endgroup$ – Takeshi Gouda 23 hours ago

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