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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.

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.

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.

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?

$\textbf{Comments:}$ 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.

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.

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.

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.

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?

$\textbf{Comments:}$ 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 $$ N_X:= \gamma_{5}^* \otimes \gamma_2^{* 2} \oplus \gamma_5^* \otimes T^*\mathbb{R P}^2 \otimes \gamma_2^* \big{|}_X, $$ 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.

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.

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?

$\textbf{Comments:}$ 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.

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.