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In an article there are the following sentences:

The euclidean group $SE(2)=\left\{\left[\begin{array}{cc}1 & 0\\v & R\end{array}\right]:v\in \mathbf{R}^{2\times1}\text{ and }R\in SO(2)\right\}$ is a real three dimensional conected matrix Lie group and its associated Lie alegbra is given by $$se(2)=\left\{\left[\begin{array}{ccc}0 & 0 & 0\\x_1 & 0 & -x_3\\x_2 & x_3 & 0\end{array}\right]:x_1,x_2,x_3\in \mathbf{R}\right\}.$$ The author identify the dual space $se(2)^*$ with matrices of the form $$\left[\begin{array}{ccc}0 & p_1 & p_2\\0 & 0 & \frac{1}{2}p_3\\0 & -\frac{1}{2}p_3 & 0\end{array}\right].$$

My question is: how did he "identify" this?

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It wouldn't hurt to give us the title and author of the article... – tomasz Dec 6 '13 at 18:07
up vote 3 down vote accepted

On, I found a description of the duality pairing between a Lie Algebra and its dual, namely $\langle X, \alpha \rangle = trace(X\alpha)$ where $X$ is in the algebra and $\alpha$ is in the dual.

But before going into that, let's talk about dual spaces a little. The dual of a vector space $V$ is the set of all linear maps from $V$ to $\mathbb R$, with addition being "addition of functions" and scalar multiplication being "multiply the function by a constant." Page 16 of gives a rather nice and concise description of the dual of a map, which amounts to this: if $T: V \rightarrow W$ is a linear transformation whose matrix with respect to some bases of $V$ and $W$ is $A$, then you can build $T^{*} : W^{*} \rightarrow V^{*}$, and build the matrix $B$ for that with respect to the dual bases of $V^{*}$ and $W^{*}$. Turns out that $B$ is just the transpose of $A$. (If $v_1, \ldots , v_n$ is a basis for $V$, the "dual basis" is $\phi_1, \phi_2, \ldots \phi_n$, where $\phi_i(v_j)$ is zero unless $i = j$, in which case it's $1$.) End of dual space discussion.

So you've got $se(3)$, consisting of a bunch of matrices, which can be regarded as representing linear maps $\mathbb R^3 \rightarrow \mathbb R^3$ with respect to the standard basis. The duals of these maps, expressed with respect to the standard dual basis, consist of the transposes of these matrices. These transposes are matrices of the form $$ \left[\begin{array}{ccc}0 & a_1 & a_2\\0 & 0 & a_3\\0 & -a_3 & 0\end{array}\right] $$

That's almost what your author wrote, except that there's the factor of $1/2$. Let's see if we can make sense of that. That's where the "trace pairing" above comes in.

If you look at matrices in $se(2)$ as you've described them above, and put in $x_1 = 1, x_2 = x_3 = 0$ to get a matrix $X_1$, you find that putting in $p_1 = 1, p_2 = p_3 = 0$ gives you a matrix $\alpha_1$ with $tr(X_1 \alpha_1) = 1$. You can similarly build $X_2, X_3$ and $\alpha_2$ and $\alpha_3$, with the property that $tr(X_i \alpha_j) = 0$ unless $i = j$, and $tr(X_i \alpha_i) = 1$, i.e., the $\alpha_j$s form a dual basis to the $X_i$s.

Once you're given the $X_i$s (as the author provided) and you know that the pairing is given by "trace-of-the-product", finding the dual basis amounts to finding some vectors (the columns of the $\alpha_j$s) that are perpendicular to certain rows certain $X_i$s so that the traces come out right, and making sure that the matrices you find are actually elements of the dual algebra (whose definition I don't have).

Let me go ahead and make this explicit. A basis for the Lie Algebra is $$ X_1 = \left[\begin{array}{ccc}0 & 0 & 0\\1 & 0 & 0\\0 & 0 & 0\end{array}\right]\\ X_2 = \left[\begin{array}{ccc}0 & 0 & 0\\0 & 0 & 0\\1 & 0 & 0\end{array}\right]\\ X_3 = \left[\begin{array}{ccc}0 & 0 & 0\\0 & 0 & -1\\0 & 1 & 0\end{array}\right] $$

To find $\alpha_1$, you're looking for a matrix, of the form I described above, whose pairings with $X_2$ and $X_3$ both yield $0$. So let's write $$ \alpha_1 = \left[\begin{array}{ccc}0 & b & c\\ 0 & 0& f\\0 &-f &0\end{array}\right]\\ $$ and compute the entries of $X_2 \alpha_1$: $$ X_2\alpha_1 = \left[\begin{array}{ccc}0& 0 & 0\\0&0& 0\\0 & b &c\end{array}\right]\\ $$ The trace of this is just $c$, so $c = 0$. Similarly, we find that $X_3 \alpha_1$ is $$ X_3\alpha_1 = \left[\begin{array}{ccc}0& 0 & 0\\0&f& 0\\0 & 0 &f\end{array}\right]\\ $$ whose trace is $2f$, from which we conclude that $f$ = 0.

Finally, look at $X_1 \alpha_1$. That's $$ X_1\alpha_1 = \left[\begin{array}{ccc}0& 0 & 0\\0&b& c\\0 & 0 &0\end{array}\right]\\ $$ whose trace is supposed to be $1$, so $b = 1$. So $\alpha_1$ must be

$$ \alpha_1 = \left[\begin{array}{ccc}0& 1 & 0\\0&0& 0\\0 &0 &0\end{array}\right]\\ $$

Essentially the same analysis shows that $\alpha_2$ must be $$ \alpha_2 = \left[\begin{array}{ccc}0& 0 & 1\\0&0& 0\\0 &0 &0\end{array}\right]\\ $$ and $\alpha_3$ must be (because this time, the trace, $2f$, must equal 1) $$ \alpha_3 = \left[\begin{array}{ccc}0& 0 & 0\\0&0& \frac{1}{2}\\0 &-\frac{1}{2} &0\end{array}\right]\\ $$

Taking the linear combination $p_1 \alpha_1 + p_2 \alpha_2 + p_3 \alpha_3$ gives the expression that your author wrote.

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thank you very much. You just give me an excellent mathematics lesson. – Martial Dec 9 '13 at 8:58

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