Real orthogonal matrices with fixed entry value has zero Haar-measure? Consider the set of real orthogonal matrices of size $n \times n$ such that one entry, say $a_{i,j}$ for fixed $i$ and $j$, satisfies $a_{i,j}=0$. Has this set zero Haar-measure? A simple proof, in affirmative (which I believe) or negative case, would be appreciated. Many Thanks in advance.
 A: We can actually show that 
$$G_{ij} = \{ A \in O(n) : a_{ij} = 0\}$$
is a smooth submanifolds in $O(n)$. In particular it has zero Haar measure. 
To see this, let $F: O(n) \to \mathbb R$ be defined by $F(A) = a_{ij}$. Then $G_{ij } = F^{-1}(0)$. To show that $G_{ij}$ is smooth, it suffices to check that $0$ is a regular value of $F$. 
Now let $A\in G_{ij}$. We have 
$$T_AO(n) = \{V \in M_n(\mathbb R): V+V^T = 0\}$$
(see the comment below) and $DF_A :T_A O(n)  \to \mathbb R$, where
$$DF_A(V) = (AV)_{ij}.$$
(Here $(\cdot)_{ij}$ is the $(i, j)$ entry of the matrix) 
Next we need to check $DF_A$ is nonzero for all $A\in G_{ij}$. Now as $A \in O(n)$, there is $j_0\neq j$ so that $a_{ij_0}\neq 0$. Now let $V$ be the matrix so that $V_{j_0 j } =1$, $V_{j j_0} = -1$ and $V_{ij} = 0$ for all other entries. Thus $V\in T_A O(n)$ and 
$$DF_A(V) = (AV)_{ij} = \sum_k a_{ik}V_{kj} = a_{i j_0} \neq 0.$$
Remark
The above construction actually shows that 
$$G_{ij, c} = \{A \in O(n): a_{ij} = c\}$$
is a submanifold of $O(n)$ when $|c|\le 1$: When $|c|<1$, the same arguement works as we are still able to pick that $j_0$. When $|c| = 1$, $G_{ij, c}$ is actually a copy of $O(n-1)$, thus is also a smooth submanifold. 
Further Remark
As $O(n)$ is a smooth Lie group, the Haar measure $\mu$ is given by a Riemannian metric. As a result, under any local coordinate $U$ of $O(n)$ with coordinate function $(x^1, \cdots, x^m)$, we have $\mu = G dx^1\cdots dx^m$, where $G$ is a positive smooth function on $U$. Thus $\mu$ and the standard Lebesgue measure $dx^1\cdots dx^m$ are absolutely continuous to each other. This shows, in particular, that any smooth submanifold $M$ in $O(n)$ with dimension strictly less than that $O(n)$ has zero Haar measure, as one can always find a local coordinate $U$ so that $U \cap M = \{x_1= \cdots x_k= 0\}$, where $k$ is the codimension of $M$. 
For your second question, in general a tangent plane of a manifold $M$ at $x$ is given by all curves $\gamma : I \to M$ so that $\gamma(0) = x$, with the identification $\gamma'(0) \in T_xM$. Now the manifold is $O(n)$ and $A\in O(n)$. Thus all curves in $O(n)$ passing through $A$ is given by $AB(t)$, where $B(t)$ is a curve in $O(n)$ passing through $I$. Now as $B^*B = I$, differentiating this with respect to $t$ and then restrict to $t=0$ gives $\dot B^* + \dot B = 0\}$. Thus 
$$T_AO(n) = \{A V: V^* + V = 0\}.$$
