"Canonical" representation of SO(3,1) $SO(3,1)$ is a subgroup of $SL(4,C)$ and thus admits a "canonical" representation on a 4-dimensional complex vector space. Is that representation irreducible?
 A: Yes, it is.  It is often called the standard representation, and will appear in any list of irreps. of $SO(3,1)$, often under this name.
A: Suppose $V$ is a nontrivial $SO(3,1)$-invariant subspace of $\mathbb C^4$, and let $(p,q,r,s)$ be a nonzero element of it.
If $p=q=r=0$, then scaling by $1/s$ we find that $(0,0,0,1)\in V$, and then by the usual arguments from the $\mathbb R^4$ case, the entire standard basis for $\mathbb C^4$ is in $V$.
Otherwise assume without loss of generality that $p\ne 0$. The set of scalar multiples of $SO(3,1)$ contains matrices arbitrarily close to $\scriptstyle\begin{pmatrix}1&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&1\end{pmatrix}$ (which is a limit of Lorentz boosts each scaled by $\frac{1}{\gamma}$), so because $V$ is closed, it must contain $(p+s,0,0,p+s)$. Scaling by $\frac 1{p+s}$, we find that $V$ contains $(1,0,0,1)$, and by the standard arguments from the real case $V$ again contains the standard basis.
This won't work if $p+s=0$, but in that case just start with $(-p,-q,r,s)$ instead of $(p,q,r,s)$.
In either case, $V=\mathbb C^4$, and since it was an arbitrary nontrivial invariant subspace, the representation is irreducible.
A: Perhaps I'm wrong but it seems to me that the corresponding Lie algebra representation $sl(2,C)\rightarrow sl(4,C)$ maps $(\begin{array}{cc}1&0\\
0&-1\end{array})$ to $(\begin{array}{cccc}0&0&0&2\\
0&0&0&0\\
0&0&0&0\\
-2&0&0&0\end{array})$ whose highest eigenvalue is 2. 
So the highest weight of this representation should be 2 while the highest weight of the irreducible representation is 3. (Representations of $sl(2,C) are classified by highest weight, which is just the highest eigenvalue of the Image of diag(1,-1).)
Where is my mistake?
