Taylor series for unitary operator in Weinberg On page 54 of Weinberg's QFT I, he says that an element $T(\theta)$ of a connected Lie group can be represented by a unitary operator $U(T(\theta))$ acting on the physical Hilbert space.  Near the identity, he says that
$$U(T(\theta)) = 1 + i\theta^a t_a + \frac{1}{2}\theta^a\theta^bt_{ab} + \ldots. \tag{2.2.17}$$
Weinberg then states that $t_a$, $t_{ab}$, ... are Hermitian.  I can see why $t_a$ must be by expanding to order $\mathcal{O}(\theta)$ and invoking unitarity.  However, expanding to $\mathcal{O}(\theta^2)$ gives
$$t_at_b = \frac{1}{2}(t_{ab} + t^\dagger_{ab})\tag{2},$$
so it seems the same reasoning cannot be used to show that $t_{ab}$ is Hermitian.  Why, then, is it?
 A: *

*OP got a good point. In the expansion
$$ \begin{align}U(T(\theta)) ~=~& {\bf 1} + i\theta^a t_a + \frac{1}{2}\theta^a\theta^b t_{ab} + {\cal O}(\theta^3), \cr 
&\theta^a \in \mathbb{R},\qquad t_{ab}~=~t_{ba}, \end{align}\tag{2.2.17} $$
it is not immediately clear if $t_{ab}$ is Hermitian as Weinberg claims$^1$. In eq. (2.2.17) $U$ is an unitary representation of a Lie group $G$, whose elements $T(\theta)\in G$ are parametrized by real parameters $\theta^a$. In more detail, the group product 
$$ T(\bar{\theta})T(\theta)~=~U(f(\bar{\theta},\theta)) \tag{2.2.15} $$
is captured by real functions 
$$f^a(\bar{\theta},\theta)~=~\theta^a+\bar{\theta}^a + f^a{}_{bc} \bar{\theta}^b\theta^c+\ldots.\tag{2.2.19}$$ 
This leads to
$$ t_{bc}~=~-t_bt_c -i t_a f^a{}_{bc}. \tag{2.2.21}$$
The symmetric combination is
$$ 2t_{bc}~=~-\{t_b,t_c\}_+ -i t_a \left(f^a{}_{bc}+f^a{}_{cb}\right), $$
so $t_{bc}$ is Hermitian iff the last term vanishes, cf eq. (1') below.

*OP's last eq. (2) is not correct. From eq. (2.2.17), the unitarity condition 
$$U^{\dagger}U~=~{\bf 1}~=~UU^{\dagger}$$
yields to second order in $\theta$ that
$$ t^{\dagger}_a~=~t_a, \tag{1'} $$
and
$$ t^{\dagger}_{ab}+t_{ab}+\{t_a,t_b\}_+~=~0. \tag{2'}$$ 
References:


*

*S. Weinberg, Quantum Theory of Fields, Vol. 1, 1995; eq. (2.2.17).


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$^1$ One may speculate that Weinberg implicitly assumes that $T(-\theta)=T(\theta)^{-1}$ so that $U(T(-\theta))=U(T(\theta))^{\dagger}$, which implies that $t_{ab}$ is indeed Hermitian.
