Download Math 215A HW3 Solution to Problem 1

Math 215A HW3 Solution to Problem 1
Brandon Van Over
March 2017
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Remark
The connectedness arguments are outlined in chapter 1 section 15 of Bredon.
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Showing Smoothness
We note that in multiplying two matrices, the entries of the product are polynomials of the
entries of the two matrices, and is smooth. We also note that Cramer’s rule gives us that the
inverse of a matrix is given by a matrix with rational functions ( for which the denominator
does not vanish!) which are the entries of the originial matrix, which is clearly smooth.
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Showing they are manifolds and Calculating Dimension
We first note that the determinant is a polynomial of the entries of the matrix and hence it
is a smooth function. Given that GLn (R) is the inverse image of the open set R \ {0}, we
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have that GLn (R) is an open subset of Rn which is clearly a submanifold of Rn , and hence
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a submanifold. An open subset of Rn has dimension n2 , and hence GLn (R) has dimension
n2 .
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We show that any nonzero x ∈ R is a regular value of det : Rn → R. From linear
algebra we know that we can expand the determinant along any row or column and yield
the
result, so we expand our determinant along the i-th row to obtain det(X) =
Pn same i+j
aij mij where mij is the value of the determinant of the i, j-th minor of X.
j=1 (−1)
Taking the partial with respect to aij yields
∂det
= (−1)i+j mij
∂aij
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Now since this is a map from Rn → R, and Tx (Rn ) = Rn , our Jacobian is given by a 1 × n2
matrix with entries being the (−1)i+j mij . Now given that we are considering the matrices
1
which have nonzero determinant, we must have that at least one mij is not zero, which gives
our map rank at least one, and hence any nonzero value of R is a regular value of det.
By the regular value theorem, SLn (R) = det−1 ({1}) is a submanifold of dimension n2 − 1 of
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Rn .
To show that O(n) is a manifold, we consider the map φ : M (n) → Sym(n) defined by
A → AAt where Sym(n) is the space of n × n symmetric matrices, i.e. those such that
At = A, and show that I is a regular value. We note that this space has dimension n(n+1)
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(the diagonal must be fixed and once you choose the everything above the diagonal you have
chosen everything below it), hence in showing that I is regular value we will have established
that O(n) is a submanifold of dimension n2 − n(n+1)
= n(n−1)
. Let A ∈ O(n) = φ−1 (I). The
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2
(A)
derivative of this map is given by f (A+tB)−f
as t → 0, which gives us
t
(A + tB)(A + tB)t − AAt
AAt + tAB t + tBAt + t2 BB t − AAt
=
= AB t + BAt
t
t
.
,
We note that this map is surjective as any C ∈ Tf (A) Sym(n) = Sym(n) has preimage CA
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and hence this map is a submersion and by the regular value theorem this is a submanifold
of dimension n(n−1)
.
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The fact that SOn (R) is a manifold of dimension n(n−1)
follows from the fact that it forms
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one of the 2 connected components of O(n), as we will see in the next section.
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Showing connectedness
In this section we make use of the following lemmata:
Lemmata 1. Let G be a topological group with a normal subgroup H. Then if H and G/H
are connected, then so is G.
Proof. Suppose that G is not connected,
i.e. G = M ∪ N where M, N are open and M ∩ N =
S
∅. Now it is clear that G = g∈G gH, and since gH is connnected by continuity of the
multiplication operation of G, we must have that it lies in either M or N for each g ∈ G.
Now consider the canonical projection mapping π : G → G/H, which is an open mapping,
and note that this gives us that G/H = π(M ) ∪ π(N ) which cannot be disjoint since G/H
is connected, but this implies that π −1 (π(M ) ∩ π(N )) 6= ∅, a contradiction.
We note that GLn (R) has at least 2 connected components, namely det−1 (0, ∞) and
det (0, −∞). We first show that the matrices of positive determinant, which we will denote
GLn (R)+ are connected by induction. We see that for the case that n = 1, the set R+ is
clearly connected, and by induction we may assume that GLk (R)+ is connected for all k < n.
Now for n > 1, we consider the action of GLn (R)+ on Rn \ {0} f : GLn (R)+ → Rn \ {0}
by A 7→ Ax, and consider this action when restricted to x0 = (0, 0, ..., 1) ∈ Rn :
−1
We note that this action is transitive, and we can see that the kernel of this map consists of
matrices of the following form:
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
x11
x21
..
.
x12
x22
..
.
x13
x23
..
.
...
...
..
.





x(n−1)1 x(n−1)2 x(n−1)3 . . .
xn1
xn2
xn3
...
x1(n−1)
x2(n−1)
..
.
x(n−1)(n−1)
xn(n−1)

0
0

.. 
.

0
1
.
which we may identify with elements of GLn−1 (R)+ × Rn−1 via the canonical inclusion.
By induction hypothesis GLn−1 (R) is connected and so is Rn−1 , and therefore their product is
connected. Now given that GLn−1 (R)+ ×Rn−1 is the kernel of the map, and matrix multiplication is continuous, we obtain a continuous bijection f : GLn (R)+ /(GLn−1 (R)+ × Rn−1 ) →
Rn \ {0} which has a continuous inverse, which follows from the fact that the projection map
−1
π : GLn (R) → GLn (R)+ /(GLn−1 (R)+ ×Rn−1 ) can be written as π = f ◦f . Hence this is a
homeomorphism and GLn (R)+ /(GLn−1 (R)+ × Rn−1 ) is connected. By the lemmata above,
since GLn (R)+ /(GLn−1 (R)+ ×Rn−1 ) and (GLn−1 (R)+ ×Rn−1 ) are connected so is GLn (R)+ .
This exact same argument may be applied to show that On (R)+ , SLn (R), SOn (R) are
connected and hence have one connected component. It remains to establish that the components with negative determinant of GLn (R) and On (R) are also connected.
Denote by I−n to be the identity matrix with the last entry on the diagonal made negative, and define a map − : GLn (R)+ → GLn (R)− by A → AI−n . We note here that
det(AI−n ) = det(A)det(I−n ) = −det(A). Since I−n is invertible, we have that this map
is a bijection, and continuous with a continuous inverse given that matrix multiplication
is continuous. Hence the positive component is homeomorphic to the negative components.
Therefore GLn (R) has the two connected components consisting of the matrices with positive
and negative determinant, and we may apply a similar argument to On (R).
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