Editing Pushforward (differential) (section)

=== Examples ===

==== Pushforward from multiplication on Lie groups ====
Given a [[Lie group]] <math>G</math>, we can use the multiplication map <math>m(-,-) : G\times G \to G</math> to get left multiplication <math>L_g = m(g,-)</math> and right multiplication <math>R_g = m(-,g)</math> maps <math>G \to G</math>. These maps can be used to construct left or right invariant vector fields on <math>G</math> from its tangent space at the origin <math>\mathfrak{g} = T_e G</math> (which is its associated [[Lie algebra]]). For example, given <math>X \in \mathfrak{g}</math> we get an associated vector field <math>\mathfrak{X}</math> on <math>G</math> defined by
<math display="block">\mathfrak{X}_g = (L_g)_*(X) \in T_g G</math>
for every <math>g \in G</math>. This can be readily computed using the curves definition of pushforward maps. If we have a curve
<math display="block">\gamma: (-1,1) \to G</math>
where
<math display="block">\gamma(0) = e \, , \quad \gamma'(0) = X</math>
we get
<math display="block">\begin{align}
(L_g)_*(X) &= (L_g\circ \gamma)'(0) \\
&= (g\cdot \gamma(t))'(0) \\
&= \frac{dg}{dt}\gamma(0) + g\cdot \frac{d\gamma}{dt} (0) \\
&= g \cdot \gamma'(0)
\end{align}</math>
since <math>L_g</math> is constant with respect to <math>\gamma</math>. This implies we can interpret the tangent spaces <math>T_g G</math> as <math>T_g G = g\cdot T_e G = g\cdot \mathfrak{g}</math>.

==== Pushforward for some Lie groups ====
For example, if <math>G</math> is the Heisenberg group given by matrices
<math display="block">H = \left\{
\begin{bmatrix}
1 & a & b \\
0 & 1 & c \\
0 & 0 & 1
\end{bmatrix} : a,b,c \in \mathbb{R}
\right\}</math>
it has Lie algebra given by the set of matrices
<math display="block">\mathfrak{h} = \left\{
\begin{bmatrix}
0 & a & b \\
0 & 0 & c \\
0 & 0 & 0
\end{bmatrix} : a,b,c \in \mathbb{R}
\right\}</math>
since we can find a path <math>\gamma:(-1,1) \to H</math> giving any real number in one of the upper matrix entries with <math>i < j</math> (i-th row and j-th column). Then, for
<math display="block">g = \begin{bmatrix}
1 & 2 & 3 \\
0 & 1 & 4 \\
0 & 0 & 1
\end{bmatrix}</math>
we have
<math display="block">T_gH = g\cdot \mathfrak{h} = 
\left\{
\begin{bmatrix}
0 & a & b + 2c \\
0 & 0 & c \\
0 & 0 & 0
\end{bmatrix} : a,b,c \in \mathbb{R}
\right\}</math>
which is equal to the original set of matrices. This is not always the case, for example, in the group
<math display="block">G = \left\{
\begin{bmatrix}
a & b \\
0 & 1/a
\end{bmatrix} : a,b \in \mathbb{R}, a \neq 0
\right\}</math>
we have its Lie algebra as the set of matrices
<math display="block">\mathfrak{g} = \left\{
\begin{bmatrix}
a & b \\
0 & -a
\end{bmatrix} : a,b \in \mathbb{R}
\right\}</math>
hence for some matrix
<math display="block">g = \begin{bmatrix}
2 & 3 \\
0 & 1/2
\end{bmatrix}</math>
we have
<math display="block">T_gG = \left\{
\begin{bmatrix}
2a & 2b - 3a \\
0 & -a/2
\end{bmatrix} : a,b\in \mathbb{R}
\right\}</math>
which is not the same set of matrices.