ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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The orthogonal group is a regular level set of dimension $n (n - 1) / 2$

Example

Let $n \geq 1$ . Identify $M_{n} (R)$ with $R^{n^{2}}$ entrywise, and identify the symmetric $n \times n$ real matrices with $R^{n (n + 1) / 2}$ by listing the entries in the positions $(i, j)$ with $i \leq j$ . Under these identifications let $f : M_{n} (R) \to {Sym}_{n} (R), f (A) = A^{T} A,$ so that $f$ is a map between Euclidean spaces of dimensions $n^{2}$ and $n (n + 1) / 2$ .

Then $f$ is $C^{\infty}$ , its derivative is $D f (A) H = A^{T} H + H^{T} A$ , and $I_{n}$ is a regular value of $f$ . Consequently $O (n) = {A \in M_{n} (R) : A^{T} A = I_{n}} = f^{- 1} (I_{n})$ is a regular level set: near each of its points it is a $C^{\infty}$ graph of dimension $n^{2} - \frac{n (n + 1)}{2} = \frac{n (n - 1)}{2},$ and its tangent space at $A \in O (n)$ is $T_{A} O (n) = {A K : K \in M_{n} (R), K^{T} = - K},$ of dimension $n (n - 1) / 2$ .

At $n = 1$ the target dimension equals the source dimension, $O (1) = {1, - 1}$ , and the graph dimension is $0$ : the two points are isolated.

Facts & Assumptions

Given: A natural number $n \geq 1$ , the entrywise identifications above, and the map $f$ with components $f_{i j} (A) = \sum_{k < n} a_{k i} a_{k j}$ for $i \leq j$ .

[L1]

Each component $f_{i j}$ is a polynomial in the entries of $A$ , so its partial derivatives of every order exist and are again polynomials, hence continuous (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ). A map each of whose components is $C^{k}$ for every $k$ is $C^{\infty}$ ( $C^{k}$ Euclidean maps and diffeomorphisms), and a map whose partial derivatives exist near a point and are continuous there is totally differentiable there (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

[L2]

If $f$ is totally differentiable at $A$ , then the directional derivative $D_{H} f (A)$ exists for every $H$ and equals $D f (A) H$ (A total derivative computes every directional derivative, and its matrix is the Jacobian).

[L3]

A $C^{1}$ map is a submersion at a point when its derivative there is surjective, and a value is regular when every point of its fibre is a submersion point (Submersions and immersions between Euclidean open sets, Regular and critical points, regular and critical values, and level sets).

[L4]

Near each of its points a regular level set of a $C^{k}$ map $U \subseteq R^{m} \to R^{N}$ is a $C^{k}$ graph of dimension $m - N$ , and its tangent space at such a point is the kernel of the derivative (A regular level set is locally a $C^{k}$ graph of dimension $m - n$ , The tangent space to a regular level set).

[L5]

A linear map is injective exactly when its kernel is trivial, and for a linear map on a finite-dimensional space the dimension of the space is the sum of the dimensions of the kernel and the image (The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial, Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

Verification

technique · direct

1.1

By [L1], $f$ is $C^{\infty}$ and totally differentiable at every $A$ .

givenL1

1.2

Let $A \in f^{- 1} (I_{n})$ , so $A^{T} A = I_{n}$ . If $A x = 0$ then $x = I_{n} x = A^{T} A x = 0$ , so by [L5] the map $x \mapsto A x$ is injective and therefore, its kernel being trivial, surjective on $R^{n}$ ; hence $A$ is invertible and $A^{- 1} = A^{T}$ , so also $A A^{T} = I_{n}$ .

givenL5algebra

2.1

Fix $A, H \in M_{n} (R)$ . Then $(A + t H)^{T} (A + t H) = A^{T} A + t (A^{T} H + H^{T} A) + t^{2} H^{T} H$ , a polynomial in $t$ with matrix coefficients, so its derivative at $t = 0$ is $A^{T} H + H^{T} A$ . By [L2] this directional derivative is $D f (A) H$ , so $D f (A) H = A^{T} H + H^{T} A$ . This matrix is symmetric, as the target requires.

step 1.1givenL2algebra

3.1

Let $S$ be symmetric and put $H = \frac{1}{2} A S$ . Then $A^{T} H = \frac{1}{2} A^{T} A S = \frac{1}{2} S$ , and $H^{T} = \frac{1}{2} S A^{T}$ gives $H^{T} A = \frac{1}{2} S A^{T} A = \frac{1}{2} S$ . By step 2.1, $D f (A) H = S$ , so $D f (A)$ is surjective onto ${Sym}_{n} (R)$ .

step 2.1step 1.2algebra

4.1

By [L3], every point of $f^{- 1} (I_{n})$ is a submersion point, so $I_{n}$ is a regular value and $f^{- 1} (I_{n}) = O (n)$ is a regular level set.

step 3.1L3

5.1

By [L4] with $m = n^{2}$ and $N = n (n + 1) / 2$ , near each of its points $O (n)$ is a $C^{\infty}$ graph of dimension $n^{2} - n (n + 1) / 2 = n (n - 1) / 2$ , and $T_{A} O (n) = \ker D f (A) = {H : A^{T} H + H^{T} A = 0}$ .

step 4.1L4algebra

6.1

If $K^{T} = - K$ and $H = A K$ , then by step 1.2 $A^{T} H = K$ and $H^{T} A = K^{T} A^{T} A = K^{T} = - K$ , so $H \in \ker D f (A)$ . Conversely, if $A^{T} H + H^{T} A = 0$ , put $K = A^{T} H$ ; then $K^{T} = H^{T} A = - K$ and $A K = A A^{T} H = H$ by step 1.2. Hence $T_{A} O (n) = {A K : K^{T} = - K}$ .

step 5.1step 1.2algebra

7.1

The map $K \mapsto A K$ is linear and injective, because $A$ is invertible by step 1.2, so by [L5] its image has the dimension of its domain. A skew-symmetric matrix is determined freely by its entries strictly above the diagonal and has zero diagonal, so the skew-symmetric matrices have dimension $n (n - 1) / 2$ , and $\dim T_{A} O (n) = n (n - 1) / 2$ .

step 6.1step 1.2L5algebra

8.1

At $n = 1$ the source and target both have dimension $1$ , $f (a) = a^{2}$ , and $f^{- 1} (1) = {1, - 1}$ , on which $f^{'} (a) = 2 a \neq 0$ ; the graph dimension $n (n - 1) / 2$ is $0$ , so each point is isolated, and the skew-symmetric $1 \times 1$ matrices are ${0}$ , in agreement with step 7.1.

step 5.1step 7.1algebra∎

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Sources

L. W. Tu, An Introduction to Manifolds, Example 11.3 (the orthogonal group) (standard reference, not scraped)
J. M. Lee, Introduction to Smooth Manifolds, Section 8 (standard reference, not scraped)

The orthogonal group is a regular level set of dimension n(n−1)/2n(n−1)/2