Smooth Manifolds Proving (1, 0) Is A Regular Value For Smoothness

Hey everyone! Today, we're diving into the fascinating world of smooth manifolds. We'll tackle a classic problem: proving that a certain topological space is indeed a smooth manifold. To do this, we'll be focusing on showing that the point $(1, 0)$ is a regular value for a specific map. This is a crucial step in demonstrating the smoothness of our manifold. So, buckle up, and let's get started!

Setting the Stage: The Map F and Mn

Before we jump into the proof, let's define the key players. We're working with a map $F$ that takes pairs of vectors from $\mathbb{R}^{n+1}$ and maps them to $\mathbb{R}^{2}$. Specifically, $F$ is defined as follows:

$F: \mathbb{R}^{n+1} \times \mathbb{R}^{n+1} \to \mathbb{R}^{2}, \quad F(p, q) = (||p||^{2}, \langle p, q \rangle)$.

Here, $||p||^{2}$ represents the squared Euclidean norm of the vector $p$, and $\langle p, q \rangle$ denotes the standard inner product (dot product) of the vectors $p$ and $q$.

Now, let's introduce our manifold candidate, $M_n$. This is defined as the preimage of the point $(1, 0)$ under the map $F$:

$M_{n} := F^{-1}((1, 0)) = \{(p, q) \in \mathbb{R}^{n+1} \times \mathbb{R}^{n+1} \mid F(p, q) = (1, 0) \}$.

In simpler terms, $M_n$ consists of all pairs of vectors $(p, q)$ in $\mathbb{R}^{n+1}$ such that the squared norm of $p$ is equal to 1 (i.e., $p$ lies on the unit sphere) and the dot product of $p$ and $q$ is equal to 0 (i.e., $p$ and $q$ are orthogonal).

Understanding the Significance: Guys, why are we doing this? Well, showing that $M_n$ is a smooth manifold is a fundamental problem in differential geometry. Smooth manifolds are the bedrock upon which much of advanced geometry and topology is built. They are spaces that locally "look like" Euclidean space, allowing us to apply the tools of calculus and analysis. Think of them as generalizations of surfaces in 3D space, but in higher dimensions!

Proving that $M_n$ is a smooth manifold involves demonstrating that it satisfies certain technical conditions. One of the most common approaches is to use the regular value theorem. This theorem states that if a map between smooth manifolds has a regular value, then the preimage of that value is itself a smooth manifold. That's where our goal of showing $(1, 0)$ is a regular value comes in! This single assertion will allow us to show that $M_n$ is a smooth manifold.

The Game Plan: Differentiability and Regular Values

Our strategy is clear: We need to show two things:

1. F is differentiable: This means that the map $F$ has a well-defined derivative at every point in its domain.
2. (1, 0) is a regular value of F: This means that the derivative of $F$ is surjective (i.e., its image spans the entire target space $\mathbb{R}^{2}$) at every point in the preimage $F^{-1}((1, 0)) = M_n$.

If we can establish these two facts, the regular value theorem will guarantee that $M_n$ is a smooth manifold. Let's dive into the details of each step.

Step 1: Proving Differentiability

The differentiability of $F$ isn't too tricky to show. Remember that $F(p, q) = (||p||^{2}, \langle p, q \rangle)$. Both components of $F$ are built from basic operations (squaring, dot product) that we know are differentiable.

Let's express $||p||^{2}$ and $\langle p, q \rangle$ in terms of the components of $p$ and $q$. Suppose $p = (p_1, p_2, ..., p_{n+1})$ and $q = (q_1, q_2, ..., q_{n+1})$. Then:

$||p||^{2} = p_1^{2} + p_2^{2} + ... + p_{n+1}^{2}$

$\langle p, q \rangle = p_1 q_1 + p_2 q_2 + ... + p_{n+1} q_{n+1}$

These expressions are polynomials in the components of $p$ and $q$. Polynomials are smooth functions, meaning they have derivatives of all orders. Therefore, both $||p||^{2}$ and $\langle p, q \rangle$ are differentiable, and consequently, the map $F$ is differentiable. Awesome!

The Derivative of F: To proceed further, we'll need to compute the derivative of $F$. The derivative of a map between Euclidean spaces is a linear transformation, which can be represented by a matrix called the Jacobian matrix. Let's find the Jacobian matrix of $F$.

Let $(p, q) \in \mathbb{R}^{n+1} \times \mathbb{R}^{n+1}$. The derivative of $F$ at $(p, q)$, denoted by $dF_{(p, q)}$, is a linear map from $\mathbb{R}^{n+1} \times \mathbb{R}^{n+1}$ to $\mathbb{R}^{2}$. We can represent it as a $2 \times 2(n+1)$ matrix. To find this matrix, we need to compute the partial derivatives of the components of $F$ with respect to the components of $p$ and $q$.

Let's start with the partial derivatives of $||p||^{2}$:

$\frac{\partial}{\partial p_i} ||p||^{2} = \frac{\partial}{\partial p_i} (p_1^{2} + p_2^{2} + ... + p_{n+1}^{2}) = 2p_i$

Now, let's compute the partial derivatives of $\langle p, q \rangle$ with respect to the components of $p$ and $q$:

$\frac{\partial}{\partial p_i} \langle p, q \rangle = \frac{\partial}{\partial p_i} (p_1 q_1 + p_2 q_2 + ... + p_{n+1} q_{n+1}) = q_i$

$\frac{\partial}{\partial q_i} \langle p, q \rangle = \frac{\partial}{\partial q_i} (p_1 q_1 + p_2 q_2 + ... + p_{n+1} q_{n+1}) = p_i$

Putting it all together, the Jacobian matrix of $F$ at $(p, q)$ is:

$dF_{(p, q)} = \begin{bmatrix} 2p_1 & 2p_2 & ... & 2p_{n+1} & 0 & 0 & ... & 0 \\ q_1 & q_2 & ... & q_{n+1} & p_1 & p_2 & ... & p_{n+1} \end{bmatrix}$

This matrix represents the linear transformation $dF_{(p, q)}$. To show that $(1, 0)$ is a regular value, we need to prove that this matrix has rank 2 (i.e., its rows are linearly independent) for all $(p, q) \in M_n$.

Step 2: Proving (1, 0) is a Regular Value

This is the heart of the proof! We need to show that $dF_{(p, q)}$ has rank 2 for all $(p, q) \in M_n$. Recall that $(p, q) \in M_n$ means that $||p||^{2} = 1$ and $\langle p, q \rangle = 0$.

The rank of a matrix is the number of linearly independent rows (or columns). For a $2 \times m$ matrix, having rank 2 simply means that the two rows are not scalar multiples of each other. So, we need to show that the rows of $dF_{(p, q)}$ are linearly independent whenever $(p, q) \in M_n$.

Let's look at the rows of $dF_{(p, q)}$:

Row 1: $(2p_1, 2p_2, ..., 2p_{n+1}, 0, 0, ..., 0)$

Row 2: $(q_1, q_2, ..., q_{n+1}, p_1, p_2, ..., p_{n+1})$

Suppose, for the sake of contradiction, that these rows are linearly dependent. This means there exists a scalar $c$ such that:

$(q_1, q_2, ..., q_{n+1}, p_1, p_2, ..., p_{n+1}) = c(2p_1, 2p_2, ..., 2p_{n+1}, 0, 0, ..., 0)$

This gives us two sets of equations:

1. $q_i = 2cp_i$ for all $i = 1, 2, ..., n+1$
2. $p_i = 0$ for all $i = 1, 2, ..., n+1$

The second set of equations immediately tells us that $p = (0, 0, ..., 0)$. But this contradicts the fact that $||p||^{2} = 1$ for $(p, q) \in M_n$. Therefore, our assumption that the rows are linearly dependent must be false!

An Alternative Approach: Another way to think about this is to consider the determinant of a $2 \times 2$ submatrix of $dF_{(p, q)}$. If we can find a $2 \times 2$ submatrix with a non-zero determinant, then the rows of $dF_{(p, q)}$ are linearly independent.

Consider the submatrix formed by the first row and the $(n+2)$-th column, and the second row and the $(n+2)$-th column. This submatrix looks like this:

$\begin{bmatrix} 2p_1 & 0 \\ q_1 & p_1 \end{bmatrix}$

The determinant of this submatrix is $2p_1^{2}$. If $p_1 \neq 0$, then the determinant is non-zero, and we're done. If $p_1 = 0$, we can consider a similar submatrix using $p_i$ for some $i \neq 1$. Since $||p||^{2} = 1$, at least one $p_i$ must be non-zero, so we can always find a submatrix with a non-zero determinant. Nice!

Conclusion of Step 2: We've successfully shown that the rows of $dF_{(p, q)}$ are linearly independent for all $(p, q) \in M_n$. This means that the derivative $dF_{(p, q)}$ has rank 2, and therefore, it is surjective. Consequently, $(1, 0)$ is a regular value of $F$.

The Grand Finale: Mn is a Smooth Manifold!

We've done it, guys! We've proven that $F$ is differentiable and that $(1, 0)$ is a regular value of $F$. Now, we can invoke the powerful regular value theorem. This theorem tells us that the preimage of a regular value under a smooth map is a smooth manifold. In our case, this means that:

$M_{n} = F^{-1}((1, 0))$ is a smooth manifold.

What does this mean in practice? This result is huge! It tells us that $M_n$, the set of pairs of orthogonal vectors $(p, q)$ in $\mathbb{R}^{n+1}$ with $||p|| = 1$, has a smooth structure. We can do calculus on $M_n$, define tangent spaces, and explore its geometric properties. This opens the door to a whole world of interesting mathematics!

Further Exploration: This result is just the beginning. We can further investigate the properties of $M_n$. For example, we can determine its dimension. Since $F$ maps to $\mathbb{R}^{2}$, and we're taking the preimage of a point, the dimension of $M_n$ is given by:

$\text{dim}(M_n) = \text{dim}(\mathbb{R}^{n+1} \times \mathbb{R}^{n+1}) - \text{dim}(\mathbb{R}^{2}) = 2(n+1) - 2 = 2n$

So, $M_n$ is a smooth manifold of dimension $2n$. We can also explore its topological properties, such as its connectedness and compactness.

Final Thoughts

Proving that a space is a smooth manifold can seem daunting at first, but by breaking it down into smaller steps and using powerful tools like the regular value theorem, we can conquer these challenges. Guys, this whole journey has been an excellent demonstration of how differential geometry connects concepts like differentiability, regular values, and the structure of manifolds. Keep exploring, keep questioning, and keep pushing the boundaries of your mathematical understanding! You've got this!