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Suppose that $V$ is a $5$ dimensional representation (with generators $\{y_1, ... , y_5\}$ of the lie group $\mathcal{g}$, with the lie algebra homomorphism $\rho: \mathcal{g} \rightarrow \mathcal{gl}(V)$. So let's say that $x$ is sent to $\rho_x$. Then $\rho_x$ is an $5 \times 5$ matrix, right? Let's say $$\rho_x = \begin{bmatrix} a_{11} & \cdots & a_{15} \\ \vdots & & \vdots\\a_{51} & \cdots & a_{55} \end{bmatrix}$$

I am wondering what the matrices for a $3$ dimensional subrepresentation $W$ (let's say that $W$ is generated by $\{y_1, y_2, y_3\}$) would look like. So we would have $\bar{\rho}: \mathcal{g} \rightarrow \mathcal{gl}(W)$ and $x$ would be sent to $\bar{\rho}_x$. This would mean that

$$\bar{\rho}_x = \begin{bmatrix} a_{11} & \cdots & a_{13} \\ \vdots & & \vdots\\a_{31} & \cdots & a_{33} \end{bmatrix}$$

Similarly, the matrix for the quotient representation $V/W$ would be $$\begin{bmatrix} a_{44} & a_{45} \\ a_{54} & a_{55} \end{bmatrix}.$$

Is that right? This might be a simple question, but I just want to know if I have the right picture in my head.

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  • $\begingroup$ Yes and no. Yes, if you can truly find a basis of $V$ so that for all $x\in\mathfrak{g}$, $\rho_x$ is a 5x5 matrix with $(\rho_x)_{ij}=0$ for $i=1,2,3$ and $j=4,5$. No, if you can't. $\endgroup$ – Aaron Oct 9 '14 at 0:01
  • $\begingroup$ @Aaron What do you mean? Isn't there any way to relate $\rho_x$ with $\bar{\rho}_x$? $\endgroup$ – Artus Oct 9 '14 at 0:26
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If $W = \text{span}\{y_{1}, y_{2}, y_{3}\}$ is in fact an invariant subspace, then for each element $g$ of your Lie algebra, $\rho(g)$ must map the the first three columns of your matrix to column vectors with $0$'s in the last two rows. Conversely, if every $\rho(g)$ is of this form, then $W$ is an invariant subspace.

If the above is true, then the matrices corresponding to the representation $W$ and $V/W$ are exactly what you wrote. What Aaron was saying is that the above might not be true.

However, note that you can always find a basis in which $W$ is the span of the first three elements if $W$ is a $3$-dimensional invariant subspace (just take a basis for $W$ and extend to a basis for $V$). So, in this basis, your representations have the given form.

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  • $\begingroup$ Thanks a lot, but isn't a subrepresentation by definition an invariant subspace? $\endgroup$ – Artus Oct 9 '14 at 10:39
  • $\begingroup$ @Artos: Yes, but there is no guarantee that you can pick a subset of a chosen basis of $V$ to give a subrepresentation, even if a proper subrepresentation exists. An example, the regular representation of the cyclic group of order 2, over $\mathbb{C}$ (not Lie algebra, but group algebra, still serve the purpose) has a basis $1,\sigma$ (corresponding to elements of the group. But can't find a subset of this basis to form a subrepresentation, but there are two 1-dimensional subrepresentations: $\mathbb{C}(1+\sigma)/2$ and $\mathbb{C}(1-\sigma)/2$. $\endgroup$ – Aaron Oct 9 '14 at 20:22

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