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To find the Lie algebra of the Heisenberg group $H$, which we know to consist of upper triangular matrices, we see that exponentials of all strictly upper triangular matrices are in $H$. I do not get the following could you explain?

"On the other hand if $X$ is any matrix such that $e^{tX}$ is upper triangular, then all the entries of $X=\frac{d}{dt}|_{t=0}e^{tX}$ which are on or below diagonal must be zero so that $X$ is upper triangular"

Thus the Lie algebra of the Heisenberg group is the space of all $3\times 3$ real matrices which are strictly upper triangular.

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    I've attempted to correct the English (please check if it preserved the intented meaning). Indeed a mention of entries $1$ on the diagonal still seems missing.2012-09-18

2 Answers 2

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Suppose you know that $e^{tX}$ is in the Heisenberg group for all $t$. Then this means that

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

where $a(t),b(t),c(t)$ are smooth functions. Now if you take the derivative of $e^{tX}$ at $t = 0$, you are taking the derivative of the individual components of the matrix on the right. This means that

$\frac{d}{dt}e^{tX}\bigg|_{t = 0} = X =\left(\begin{array}{ccc} \frac{d}{dt}1 & a'(0) & b'(0) \\ 0 & \frac{d}{dt}1 & c'(0) \\ 0 & 0 & \frac{d}{dt}1 \end{array}\right) = \left(\begin{array}{ccc} 0 & a'(0) & b'(0) \\ 0 & 0 & c'(0) \\ 0 & 0 & 0\end{array}\right).$

$a'(0), b'(0)$ and $c'(0)$ are just real numbers and the calculation above shows that $X$ is strictly upper triangular.

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The statement you cite is not true, but can be made so by adding "for all $t$" and "with entries $1$ on the diagonal" to the hypothesis. The examples of $\begin{pmatrix}0&1&0\\-1&0&0\\0&0&0\end{pmatrix}$ with $t=2\pi$, respectively of any diagonal matrix, show the necessity of these additions.