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Changed the question to whether $\tilde{H}_p^*$ intersects transversally.
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Ritwik
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Let $\mathcal{D} \approx \mathbb{P}^{\delta_d}$ be the space of homogeneous degree $d$ polynomials in three vriables, where $\delta_d = \frac{d(d+3)}{2}$. Let $$ X \subset \mathcal{D} \times \mathbb{P}^2$$

be a smooth embedded complex submanifold, not necessarily closed. Given a point $p\in \mathbb{P}^2$, we get a hyperplane $$\tilde{H}_p \in \mathcal{D} \times \mathbb{P}^2.$$ Note that a point $p$ first of all gives a hyperplane $H_p$ in $\mathcal{D}$ (which is the space of degree $d$ polynomials passing through the point $p$). This gives us a hyperplane $$ \tilde{H}_p := H_p\times \mathbb{P}^2 \in \mathcal{D} \times \mathbb{P}^2.$$ Is it true that ifLet us further define $\tilde{H}_p$ is not transverse$H_p^* \subset H_p$ to $X$, then for every open neighborhood $U$ of $p$ in $\mathbb{P}^2$, there exists abe the pointspace of degree $q\in U$$d$ curves such that the hyperplane $\tilde{H}_q$ does intersect $X$ transversally? This seems to intuitively say that if something$p$ is not transversea smooth point of the then you can ``perturb'' it so that it is transversecurve. Similarly define More precisely, I want to claim$$ \tilde{H}_p^* := H_p^*\times \mathbb{P}^2 \in \mathcal{D} \times \mathbb{P}^2.$$

Is it true that the setfor almost all choices of points $q$ where $\tilde{H}_q$ intersects $X$ transversally is an open$p \in \mathbb{P}^2$, dense subset of $\mathbb{P}^2$. It is certainly open. Please note that apriori it$\tilde{H}^*_p$ is possible that there does not exist any point $q$ such thattransverse to $\tilde{H}_q$ intersects $X$ transversally. This is the part I don't see how to prove, although intuitively it seems obvious.?

Let $\mathcal{D} \approx \mathbb{P}^{\delta_d}$ be the space of homogeneous degree $d$ polynomials in three vriables, where $\delta_d = \frac{d(d+3)}{2}$. Let $$ X \subset \mathcal{D} \times \mathbb{P}^2$$

be a smooth embedded complex submanifold, not necessarily closed. Given a point $p\in \mathbb{P}^2$, we get a hyperplane $$\tilde{H}_p \in \mathcal{D} \times \mathbb{P}^2.$$ Note that a point $p$ first of all gives a hyperplane $H_p$ in $\mathcal{D}$ (which is the space of degree $d$ polynomials passing through the point $p$). This gives us a hyperplane $$ \tilde{H}_p := H_p\times \mathbb{P}^2 \in \mathcal{D} \times \mathbb{P}^2.$$ Is it true that if $\tilde{H}_p$ is not transverse to $X$, then for every open neighborhood $U$ of $p$ in $\mathbb{P}^2$, there exists a point $q\in U$ such that the hyperplane $\tilde{H}_q$ does intersect $X$ transversally? This seems to intuitively say that if something is not transverse then you can ``perturb'' it so that it is transverse. More precisely, I want to claim that the set of points $q$ where $\tilde{H}_q$ intersects $X$ transversally is an open, dense subset of $\mathbb{P}^2$. It is certainly open. Please note that apriori it is possible that there does not exist any point $q$ such that $\tilde{H}_q$ intersects $X$ transversally. This is the part I don't see how to prove, although intuitively it seems obvious.

Let $\mathcal{D} \approx \mathbb{P}^{\delta_d}$ be the space of homogeneous degree $d$ polynomials in three vriables, where $\delta_d = \frac{d(d+3)}{2}$. Let $$ X \subset \mathcal{D} \times \mathbb{P}^2$$

be a smooth embedded complex submanifold, not necessarily closed. Given a point $p\in \mathbb{P}^2$, we get a hyperplane $$\tilde{H}_p \in \mathcal{D} \times \mathbb{P}^2.$$ Note that a point $p$ first of all gives a hyperplane $H_p$ in $\mathcal{D}$ (which is the space of degree $d$ polynomials passing through the point $p$). This gives us a hyperplane $$ \tilde{H}_p := H_p\times \mathbb{P}^2 \in \mathcal{D} \times \mathbb{P}^2.$$ Let us further define $H_p^* \subset H_p$ to be the space of degree $d$ curves such that $p$ is a smooth point of the curve. Similarly define $$ \tilde{H}_p^* := H_p^*\times \mathbb{P}^2 \in \mathcal{D} \times \mathbb{P}^2.$$

Is it true that for almost all choices of $p \in \mathbb{P}^2$, $\tilde{H}^*_p$ is transverse to $X$?

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Ritwik
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Let $\mathcal{D} \approx \mathbb{P}^{\delta_d}$ be the space of homogeneous degree $d$ polynomials in three vriables, where $\delta_d = \frac{d(d+3)}{2}$. Let $$ X \subset \mathcal{D} \times \mathbb{P}^2$$

be a smooth embedded complex submanifold, not necessarily closed. Given a point $p\in \mathbb{P}^2$, we get a hyperplane $$\tilde{H}_p \in \mathcal{D} \times \mathbb{P}^2.$$ Note that a point $p$ first of all gives a hyperplane $H_p$ in $\mathcal{D}$ (which is the space of degree $d$ polynomials passing through the point $p$). This gives us a hyperplane in $$ \tilde{H}_p := H_p\times \mathbb{P}^2 \in \mathcal{D} \times \mathbb{P}^2.$$ Is it true that if $\tilde{H}_p$ is not transverse to $X$, then there exists an open neighborhoodfor every open neighborhood $U$ of $p$ in $\mathbb{P^2}$$\mathbb{P}^2$, such thatthere exists a for all pointspoint $q\in U$ such that are not equal to $p$, the hyperplane $\tilde{H}_q$ does intersect $X$ transversally? This seems to intuitively say that if something is not transverse then you can ``perturb'' it so that it is transverse. Is More precisely, I want to claim that the set of points $q$ where $\tilde{H}_q$ intersects $X$ transversally is an open, dense subset of $\mathbb{P}^2$. It is certainly open. Please note that apriori it is possible that there does some general theorem in differential topologynot exist any point $q$ such that implies$\tilde{H}_q$ intersects this immediately?$X$ transversally. This is the part I don't see how to prove, although intuitively it seems obvious.

Let $\mathcal{D} \approx \mathbb{P}^{\delta_d}$ be the space of homogeneous degree $d$ polynomials in three vriables, where $\delta_d = \frac{d(d+3)}{2}$. Let $$ X \subset \mathcal{D} \times \mathbb{P}^2$$

be a smooth embedded complex submanifold, not necessarily closed. Given a point $p\in \mathbb{P}^2$, we get a hyperplane $$\tilde{H}_p \in \mathcal{D} \times \mathbb{P}^2.$$ Note that a point $p$ first of all gives a hyperplane $H_p$ in $\mathcal{D}$ (which is the degree $d$ polynomials passing through the point $p$). This gives a hyperplane in $$ \tilde{H}_p := H_p\times \mathbb{P}^2 \in \mathcal{D} \times \mathbb{P}^2.$$ Is it true that if $\tilde{H}_p$ is not transverse to $X$, then there exists an open neighborhood $U$ of $p$ in $\mathbb{P^2}$, such that for all points $q\in U$ that are not equal to $p$, the hyperplane $\tilde{H}_q$ does intersect $X$ transversally? This seems to intuitively say that if something is not transverse then you can ``perturb'' it so that it is transverse. Is there some general theorem in differential topology that implies this immediately?

Let $\mathcal{D} \approx \mathbb{P}^{\delta_d}$ be the space of homogeneous degree $d$ polynomials in three vriables, where $\delta_d = \frac{d(d+3)}{2}$. Let $$ X \subset \mathcal{D} \times \mathbb{P}^2$$

be a smooth embedded complex submanifold, not necessarily closed. Given a point $p\in \mathbb{P}^2$, we get a hyperplane $$\tilde{H}_p \in \mathcal{D} \times \mathbb{P}^2.$$ Note that a point $p$ first of all gives a hyperplane $H_p$ in $\mathcal{D}$ (which is the space of degree $d$ polynomials passing through the point $p$). This gives us a hyperplane $$ \tilde{H}_p := H_p\times \mathbb{P}^2 \in \mathcal{D} \times \mathbb{P}^2.$$ Is it true that if $\tilde{H}_p$ is not transverse to $X$, then for every open neighborhood $U$ of $p$ in $\mathbb{P}^2$, there exists a point $q\in U$ such that the hyperplane $\tilde{H}_q$ does intersect $X$ transversally? This seems to intuitively say that if something is not transverse then you can ``perturb'' it so that it is transverse. More precisely, I want to claim that the set of points $q$ where $\tilde{H}_q$ intersects $X$ transversally is an open, dense subset of $\mathbb{P}^2$. It is certainly open. Please note that apriori it is possible that there does not exist any point $q$ such that $\tilde{H}_q$ intersects $X$ transversally. This is the part I don't see how to prove, although intuitively it seems obvious.

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Ritwik
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Can you ``perturb'' a submanifold to intersect transversally with any other smooth submanifold of projective space?

Let $\mathcal{D} \approx \mathbb{P}^{\delta_d}$ be the space of homogeneous degree $d$ polynomials in three vriables, where $\delta_d = \frac{d(d+3)}{2}$. Let $$ X \subset \mathcal{D} \times \mathbb{P}^2$$

be a smooth embedded complex submanifold, not necessarily closed. Given a point $p\in \mathbb{P}^2$, we get a hyperplane $$\tilde{H}_p \in \mathcal{D} \times \mathbb{P}^2.$$ Note that a point $p$ first of all gives a hyperplane $H_p$ in $\mathcal{D}$ (which is the degree $d$ polynomials passing through the point $p$). This gives a hyperplane in $$ \tilde{H}_p := H_p\times \mathbb{P}^2 \in \mathcal{D} \times \mathbb{P}^2.$$ Is it true that if $\tilde{H}_p$ is not transverse to $X$, then there exists an open neighborhood $U$ of $p$ in $\mathbb{P^2}$, such that for all points $q\in U$ that are not equal to $p$, the hyperplane $\tilde{H}_q$ does intersect $X$ transversally? This seems to intuitively say that if something is not transverse then you can ``perturb'' it so that it is transverse. Is there some general theorem in differential topology that implies this immediately?