323: Intersection of 2 Transversal Regular Submanifolds of C^\infty Manifold Is Regular Submanifold of Specific Codimension

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that intersection of 2 transversal regular submanifolds of $C^{\mathrm{\infty }}$ manifold is regular submanifold of specific codimension

---

Topics

About: $C^{\mathrm{\infty }}$ manifold

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

---

Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ manifold.
• The reader knows a definition of regular submanifold.
• The reader knows a definition of transversal regular submanifolds.
• The reader admits the transversality theorem: for any transversal map, the preimage of the regular submanifold on the codomain is a regular submanifold of codimension of the codimension of the regular submanifold on the codomain.
• The reader admits the proposition that any regular submanifold of any regular submanifold of any $C^{\mathrm{\infty }}$ manifold is a regular submanifold of the base manifold, of codimension of the codimension of the child submanifold plus the codimension of the grandchild submanifold with respect to the child submanifold.

---

Target Context

• The reader will have a description and a proof of the proposition that the intersection of any 2 transversal regular submanifolds of any $C^{\mathrm{\infty }}$ manifold is a regular submanifold of codimension of the sum of the codimensions of the regular submanifolds.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Description

For any $C^{\mathrm{\infty }}$ manifold, M, and any transversal regular submanifolds, $M_1,M_2\subseteq M$, the intersection of the transversal regular submanifolds, $M_1\cap M_2$, is a regular submanifold of $M$ of codimension of the sum of the codimensions of $M_1$ and $M_2$.

---

2: Proof

Think of the inclusion map, $i:M_1\to M$, which is $C^{\mathrm{\infty }}$ because $M_1$ has the adapting charts and $i$ with respect to any adapting chart and the adapted chart is obviously $C^{\mathrm{\infty }}$. $i$ is a transversal map from $M_1$ to $M_2$, because for each $p\in i^{-1}(M_2)$, $i_{\ast }(T_p{M}_1)+i_{\ast }^{\prime }(T_{i(p)}{M}_2)=T_{i(p)}M$ where $i^{\prime }$ is the inclusion map, $i^{\prime }:M_2\to M$, because as $i$ is an inclusion, $i(p)=p$, so, that equation equals $i_{\ast }(T_p{M}_1)+i_{\ast }^{\prime }(T_p{M}_2)=T_{p}M$, which is nothing but the definition of the transversality of $M_1$ and $M_2$ (although the definition is prevalently expressed like "$T_p{M}_1+T_p{M}_2=T_{p}M$", it is really a sloppy expression because any vector on $T_p{M}_1$ cannot be added to any vector on $T_p{M}_2$ because they are on different spaces; the vectors can be added only because they are pushed forward to the same $T_{p}M$ space). By the transversality theorem: for any transversal map, the preimage of the regular submanifold on the codomain is a regular submanifold of codimension of the codimension of the regular submanifold on the codomain, $i^{-1}(M_2)=M_1\cap M_2$ is a regular submanifold of $M_1$ of codimension of the codimension of $M_2$ (with respect to $M$), denoted as $codim(M_2,M)$. By the proposition that any regular submanifold of any regular submanifold of any $C^{\mathrm{\infty }}$ manifold is a regular submanifold of the base manifold, of codimension of the codimension of the child submanifold plus the codimension of the grandchild submanifold with respect to the child submanifold, $M_1\cap M_2$ is a regular submanifold of $M$ of codimension of $codim(M1,M)+codim(M2,M)$.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>