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Let H$_{3}$(R) denote the Heisenberg group of 3 x 3 real matrices. For any A $\in$ H$_{3}$(R) such that

A = $\left( \begin{array}{ccc} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{array} \right)$ , $\exists$ a continuous path A(t), 0 $\leq$ t $\leq$ 1, from I to A given by

A(t) = $ \left( \begin{array}{ccc} 1 & a(t) & b(t) \\ 0 & 1 & c(t) \\ 0 & 0 & 1 \end{array} \right)$ .

Let v be the tangent vector of A(t) at identity (t=0), then v= $\frac{d}{dt}$ A(t)|$_{t=o}$.

Then shouldn't

v = A'(0) = $ \left( \begin{array}{ccc} 1 & \frac {d}{dt}0 & \frac{d}{dt}0 \\ 0 & 1 & \frac{d}{dt}0 \\ 0 & 0 & 1 \end{array} \right)$ = $ \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$ ?

If this is correct, then is the tangent space of H$_{3}$(R) the standard, orthonormal basis of R$^3$?

(Apologies if this is horribly fallacious, I'm new to lie algebra)

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2 Answers 2

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The notation $A'(0)$ does not mean "differentiate the constant $A(0)$." Indeed, if you differentiate a constant, you will always get $0$. (In this case, you would get the zero matrix, so even the $1$s on the diagonal wouldn't make sense. Moreover, the identity matrix is not "the standard orthonormal basis of $\mathbb{R}^3$"; its columns are, but that's utterly irrelevant. The tangent space of $H_3(\mathbb{R})$ will consist of of matrices, not of column vectors.)

Rather, $A'(0)$ means "differentiate $A(t)$, then plug in $t=0$."

So if you have a differentiable path $A(t)$, each of $a(t),b(t),c(t)$ must be differentiable, and

$$ A'(t)= \frac{d}{dt} \begin{pmatrix} 1 & a(t) & b(t) \\ 0 & 1 & c(t) \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & a'(t) & b'(t) \\ 0 & 0 & c'(t) \\ 0 & 0 & 0 \end{pmatrix}. $$

Then you plug $t=0$ into this.

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No. Choose a path $\gamma$ made up of matrices and passing through the identity; formally, $\gamma : (-r,r) \to H_3 (\Bbb R)$ with $\gamma (0) = 1$ (the $3 \times 3$ identity matrix, and some $r>0$). Writing in components, $\gamma (t) = \begin{pmatrix} 1 & a(t) & b(t) \\ 0 & 1 & c(t) \\ 0 & 0 & 1 \end{pmatrix}$. Now, the tangent space in $1$ at $H_3 (\Bbb R)$ is made up of the tangent vectors at $t=0$ of all these curves, i.e. the vectors $\gamma ' (0) = \begin{pmatrix} 0 & a'(0) & b'(0) \\ 0 & 0 & c'(0) \\ 0 & 0 & 0 \end{pmatrix}$. Since $a$, $b$ and $c$ are arbitrary, their derivatives in $0$ will be arbitrary too. Therefore, calling them $u,v,w$, it follows that the tangent space that you are after is made up of the matrices $\begin{pmatrix} 0 & u & v \\ 0 & 0 & w \\ 0 & 0 & 0 \end{pmatrix}$ with $u,v,w \in \Bbb R$.

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  • $\begingroup$ Thank you. Out of curiosity, I understood the functions a(t), b(t), c(t) to be multiplication of a constant (a, b, c) by the variable t. How is it that the derivative of these functions produces an arbitrary real number? $\endgroup$
    – user413923
    Mar 17, 2017 at 22:30
  • $\begingroup$ @Garaidh: No, the functions $a$, $b$ and $c$ are really arbitrary, because there is nothing said about them in the definition of $H_3 (\Bbb R)$. It follows that their derivatives will also be arbitrary. True, you may choose, in particular, $a(t) = tu$, $b(t) = tv$ and $c(t) = tw$, with $u,v,w,$ arbitrary - a thing which leads straight to the formula given by me. You won't be able to this in abstract Lie groups, anymore, but this is not a concern for you right now. $\endgroup$
    – Alex M.
    Mar 17, 2017 at 22:33
  • $\begingroup$ @Garaidh: Well, $a$, $b$ and $c$ are arbitrary but with the constraint that $a(0) = b(0) = c(0) = 0$, I had forgotten about this. Anyway, this changes nothing about my answer. $\endgroup$
    – Alex M.
    Mar 17, 2017 at 22:36
  • $\begingroup$ Ah. That makes sense. Thank you very much. $\endgroup$
    – user413923
    Mar 17, 2017 at 22:40

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