The electroweak Lagrangian before symmetry breaking is defined as:
$$L_{ew} = L_g + L_f + L_h + L_y$$
where the $L_g$ portion concerns 4 identically-structured vector fields
$$\vec B = [B^0,B^1,B^2,B^3]\\ \vec W_0 = [W_0^0,W_0^1,W_0^2,W_0^3]\\ \vec W_1 = [W_1^0,W_1^1,W_1^2,W_1^3]\\ \vec W_2 = [W_2^0,W_2^1,W_2^2,W_2^3]$$
By definition,
$$L_g = -\frac{1}{4}(B^{\mu\nu}B_{\mu\nu}) -\frac{1}{4}tr(W_0^{\mu\nu}W^0_{\mu\nu})-\frac{1}{4}tr(W_1^{\mu\nu}W^1_{\mu\nu})-\frac{1}{4}tr(W_2^{\mu\nu}W^2_{\mu\nu})$$
where
$$B^{ij} = \frac{dB^j}{dx^i} - \frac{dB^i}{dx^j}$$
and $W_i^{\mu\nu}$ is a $4x4$ matricx whose elements are defined by
$$W^{ab}_i=\frac{d}{dx^a}W^b_i - \frac{d}{dx^b}W^a_i - igW^a_iW^b_i + igW^b_iW^a_i$$
The lower-index $W^0_{\mu\nu}$ (and $B_{\mu\nu}$) are computed using the relation
$$W_{ab} = \sum_{i,j}^4 \eta_{ib}W^{ji}\eta_{aj}$$
where $\eta$ is the Minkowski metric
$$\eta = \begin{bmatrix} 1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}$$
I am unclear about which $\eta$ metric is to be used here, as there are 2 possible $\eta$, the alternative being
$$\eta^* = \begin{bmatrix} -1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}$$
Does it matter which version of $\eta$ or $\eta^*$ is used in the computation of $L_g$, in the scheme of the overall $L_{ew}$ computation?
Is any other metric other than $\eta$ allowed in the Lagrangian?
The computational result of
$$W_{ab}=\sum_{i,j}^4\eta_{ib}W^{ji}\eta_{aj}$$
is naturally invariant whether (+,-,-,-) or (-,+,+,+) is used as $\eta$, ie.
lowered using (+,-,-,-) gives us
and likewise using (+,-,-,-) gives us the identical
Thus, there is no operational significance to which $\eta$ is adopted for purposes of index-lowering a tensor.
The general component-wise result of $\sum_{i,j}^4 W_{ij}W^{ij}$ is thus the matrix multiplication yielding