Is the space of real conics with a singular point an orientable manifold?

Question

Consider the space of non zero real homogeneous degree $2$ polynomials in three variables upto scaling. This space is $\mathbb{R} \mathbb{P}^5$. The zero set of such a polynomial gives a real curve in $\mathbb{R} \mathbb{P}^2$. Let me define the space $X$ to be the space of real curves $[f]$ and a marked point $p$, such that the curve has a singularity at $p$, i.e. $$ X := \{ ([f], p) \in \mathbb{R} \mathbb{P}^5 \times \mathbb{R} \mathbb{P}^2: f(p) =0, ~~\nabla f|_p =0 \}. $$

$\textbf{Question $1$:}$ Is $X$ a smooth manifold of the expected dimension (four)?

$\textbf{Question $2$:}$ If yes, is $X$ orientable or non orientable?

$\textbf{Question $3$:} $ Is it obvious that $X \rightarrow \mathbb{RP}^2$ is a fiber bundle (i.e. locally trivial)?

$\textbf{Comments on Question $3$:}$ Suppose we write a homogeneous polynomial explicitly as $$ f(X,Y,Z) := A_1 X^2 + A_2 Y^2 + A_3 Z^3 + A_4 X Y + A_5 X Z + A_6 YZ.$$ Suppose we want to trivialize the bundle near the point $[0,0,1]$. It is easy to see that if we ask for $f$ to have a singularity at $[0,0,1]$ then $A_3, ~A_5, ~A_6 =0$. Let $\mathcal{U}_{\epsilon_1, \epsilon_2}$ be an open set consisting of points of the form $[\epsilon_1, \epsilon_2,1]$. It is easy to see that the map $$ h: \pi^{-1} (\mathcal{U}_{\epsilon_1, \epsilon_2}) \rightarrow \mathcal{U}_{\epsilon_1, \epsilon_2} \times \mathbb{RP}^2 $$ given by $$ h([A_1, \ldots, A_6]; p):= (p, [A_1, A_2, A_4]$$ is not a trivialization.

Of course, that doesn't prove anything; but I am just wondering if $\pi:X \rightarrow \mathbb{RP}^2$ is a fiber bundle?

$\textbf{Comments:}$ I have two arguments giving me contradictory answers. First of all, note that (assuming $X$ is a manifold), the normal bundle to $X$ in $\mathbb{R} \mathbb{P}^5 \times \mathbb{R} \mathbb{P}^2$ is given by $$ N_X:= \gamma_{5}^* \otimes \gamma_2^{* 2} \oplus \gamma_5^* \otimes T^*\mathbb{R P}^2 \otimes \gamma_2^{*2} \big{|}_X, $$ where $\gamma_n$ is the tautological line bundle over $\mathbb{RP}^n$. (I can justify this if someone is not convinced).

Note that $$ TX \approx \frac{T (\mathbb{R} \mathbb{P}^5 \times \mathbb{R} \mathbb{P}^2)}{N_X}. $$ It is now easy to see that the first stiefel whitney class of the tangent bundle of $X$ is zero, i.e. $w_1(TX) =0$. Hence $X$ is orientable.

$\textbf{The above statement is incorrect:}$ One can check that $w_1(TX) \neq 0$, which is consistent with the fact that $X$ is non-orientable.

On the other hand it seems to me that $~\pi: X \rightarrow \mathbb{RP}^2$ is a fibre bundle, with fibres $\mathbb{RP}^2$. One can show (using spectral sequences) that any $\mathbb{RP}^2$ bundle over $\mathbb{RP}^2$ is non orientable.

$\textbf{Proof of why $X$ is a manifold:} $ Let us take $\mathbb{R}^6$ to be the space of real polynomials in two variables of degree at most $2$. Hence we can write such an element $\rho \in \mathbb{R}^6$ as $$ \rho(x,y) = \rho_{00} + \rho_{10} x + \rho_{01} y + \frac{\rho_{20}}{2} x^2 + \rho_{11} x y + \frac{\rho_{02}}{2} y^2. $$

Now consider the map $$ \psi : \mathbb{R}^6 \times \mathbb{R}^2 \rightarrow \mathbb{R}^3 $$ given by $$ \psi(\rho; x,y) := \Big( \rho_{00} + \rho_{10} x + \rho_{01} y + \frac{\rho_{20}}{2} x^2 + \rho_{11} x y + \frac{\rho_{02}}{2} y^2, \\ ~\rho_{10} + \rho_{20}x + \rho_{11} y, \\ ~\rho_{01} + \rho_{11}x + \rho_{02}y \Big). $$ It is easy to see that if $\psi(\rho,0,0) =0$, then the Jacobian matrix of $\psi(\rho,x,y)$ at $(x,y)=(0,0)$ has full rank. To see why, take the partial derivative of $\psi$ with respect to $\rho_{00}$, $\rho_{10}$ and $\rho_{01}$ and plug in $(x,y) = (0,0)$. That gives us a $3\times 3$ identity matrix. Hence $\psi$ is transverse to the zero set (it is easy to see taking $(x,y) = (0,0)$ was without any loss of generality). Hence, by Implicit Function Theorem, $\psi^{-1}(0,0,0)$ is a smooth manifold (even around double lines).

Its now easy to see that the space $X$ defined will also be a manifold; by writing the evaluation map and the vertical derivative in a coordinate chart and trivialization, it reduces to the above calculation.

Answer 1

Yes, it is a smooth manifold. No, it is not orientable.

For the first, just think geometrically, i.e., without bases: Fix a $3$-dimensional vector space $V$ and consider the homogeneous quadratic polynomials on it, which is a $6$-dimensional vector space isomorphic to $\mathsf{S}^2(V^\ast)$. Let $\mathbb{P}V$ be the projectivization of $V$, i.e., the space of $1$-dimensional subspaces of $V$. Given $L\in\mathbb{P}V$, the set of elements of $\mathsf{S}^2(V^\ast)$ that represent conics singular at $L$ is simply the $3$-dimensional space $\mathsf{S}^2(L^\perp)\subset \mathsf{S}^2(V^\ast)$, where $L^\perp\subset V^\ast$ is the space of linear functions on $V$ that vanish on $L$. In particular, the set $E\subset \mathbb{P}V\times \mathsf{S}^2(V^\ast)$ that consists of the pairs $(L,f)$ such that $f\in \mathsf{S}^2(L^\perp)$ is a smooth subbundle of rank $3$ over $\mathbb{P}V$ of the trivial bundle $\mathbb{P}V\times \mathsf{S}^2(V^\ast)$. Your space $X$ is simply $\mathbb{P}E$, the projectivization of this smooth, rank $3$ bundle. In particular, it is a smooth submanifold of $\mathbb{P}V\times \mathbb{P}\bigl(\mathsf{S}^2(V^\ast)\bigr)$ of dimension $4 = 2 + (3{-}1)$, and the projection $\mathbb{P}E\to\mathbb{P}V$ is a submersion that is a locally trivial fiber bundle.

Second, $X=\mathbb{P}E$ cannot be orientable because no projectivization of a locally trivial, rank $3$ bundle $E$ over a smooth manifold $M$ is orientable. You don't need spectral sequences to see this, you just need to produce one closed loop around which the orientation bundle is nontrivial, but these are easy to find: Just take a loop in a single fiber $\mathbb{P}E_x$ that generates that fiber's fundamental group (which is isomorphic to $\mathbb{Z}_2$). This is clearly an orientation-reversing loop in $\mathbb{P}E$.

Response to Question 3: The OP wondered why the bundle $\pi:E\to \mathbb{P}V$ is locally trivial. Here is one way to see this: Let $L_0\in \mathbb{P}V$ be a $1$-dimensional subspace and let $X,Y,Z\in V^\ast$ be a basis of the linear functions on $V$ such that $X$ and $Y$ vanish identically on $L_0$ (and hence are a basis of $(L_0)^\perp$) while $z$ does not identically vanish on $L_0$ (and hence only vanishes on $L_0$ at $0\in L_0$). Let $U\subset\mathbb{P}V$ be the open set consisting of those $L\in \mathbb{P}V$ such that $Z$ is not the zero linear functional when restricted to $L$. On $U$, there are two well-defined, smooth functions $x,y:U\to\mathbb{R}$ such that $x(L) = X(v)/Z(v)$ and $y(L) = Y(v)/Z(v)$ for some (and, hence, any) nonzero $v\in L$. Then, for each $L\in U$, the linear functions $X-x(L)\,Z$ and $Y - y(L)\,Z$ are a basis of $(L)^\perp\subset V^\ast$, and hence the quadratic forms $$ A(L) = \bigl(X-x(L)\, Z\bigr)^2,\quad B(L) = \bigl(X-x(L)\, Z\bigr)\bigl(Y-y(L)\, Z\bigr),\quad C(L) = \bigl(Y-y(L)\, Z\bigr)^2 $$ are a basis of $E_L = \mathsf{S}^2(L^\perp)$ for each $L\in U$. Thus, they define a smooth trivialization of $\pi:E\to\mathbb{P}V$ over $U$. Obviously, $\mathbb{P}V$ can be covered by such open sets $U$, and it is easy to see that the transitions on overlaps are smooth. Thus $E$ is a smooth bundle with local smooth trivializations.