The three generators of $su(2)$ satisfy the commutation relations
$$ [J_0 , J_\pm] = J_\pm , \quad [J_+, J_- ] = +2J_0 .$$
The three generators of $su(1,1)$ satisfy the commutation relations
$$ [K_0 , K_\pm] = K_\pm , \quad [K_+, K_- ] = -2K_0 .$$
Now, let us define
$$ K_0 = J_0, \; K_+ = J_+,\; K_- = - J_-. $$
It is apparent that so defined $K$'s satisfy the $su(1,1)$ algebra! Does this mean that $su(1,1)$ is actually equivalent to $su(2)$?
Where is the argument wrong?
Answer
You may indeed identify the generators in the way you did. However, the Lie algebras and Lie groups are different because – as quickly said by Qmechanic – you must use different reality conditions for the coefficients.
A general matrix in the $SU(2)$ group is written as $$ M = \exp[ i( \alpha J_+ + \bar\alpha J_- + \gamma J_0 )] $$ where $\alpha\in {\mathbb C}$ and $\gamma\in {\mathbb R}$ while the general matrix in $SU(1,1)$ is given by $$ M' = \exp [ i( \alpha_+ J_+ + \alpha_- J_- + \beta J_0 ] $$ where $\alpha_+,\alpha_-,\beta\in {\mathbb R}$ are three different real numbers.
To summarize, for $SU(2)$, the coefficients in front of $J_\pm$ are complex numbers conjugate to each other, while for $SU(1,1)$, they are two independent real numbers. (And I apologize that I am not sure whether the $i$ should be omitted in the exponent of $SU(1,1)$ only according to your convention. Probably.)
If you allow all three coefficients in front of $J_\pm,J_0$ to be three independent complex numbers, you will obtain the complexification of the group. And as Qmechanic also wrote, the complexification of both $SU(2)$ and $SU(1,1)$ is indeed the same, namely $SL(2,{\mathbb C})$.
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