Minkowski Inner Product's Shenanigans In the context of Special Relativity, so in flat spacetime, and with the metric tensor $g_{\mu \nu}$ chosen with the signature: $(+,-,-,-)$, lets consider the following four vectors:
$$x=(x_1,x_2,x_3,x_4)$$
$$y=(y_1,y_2,y_3,y_4)$$
the Minkowski inner product between the two is:
$$x^\mu y^\nu g_{\mu \nu}=x_1y_1-x_2y_2-x_3y_3-x_4y_4 \ \ \ \ \ \ \ \ \ \ (1)$$
Wonderful, but we can also interprete the presence of $g_{\mu \nu}$ as a "lowering indices agent", such that:
$$x^\mu y^\nu g_{\mu \nu}=x^\mu y_\mu$$
this has to be true, but then following the Einstein's summation convention we get:
$$x^\mu y^\nu g_{\mu \nu}=x^\mu y_\mu=x_1y_1+x_2y_2+x_3y_3+x_4y_4$$
so a different result than we previously got in $(1)$.
This is obviously absurd; where am I wrong?
 A: In your last equality, you should write $$x^\mu y_\mu=x^1y_1+x^2y_2+x^3y_3+x^4y_4\tag{1}.$$ Note that this is not equal to $x_1y_1+x_2y_2+x_3y_3+x_4y_4$. To recover your first equality you should then lower the $x$ indices. Remember $x^\mu=x_\nu g^{\mu\nu}$, so then
$$
x^1=x_1g^{11}+x_2g^{12}+x_3g^{13}+x_4g^{14}=x_1+0+0+0
$$
But
\begin{align}
x^2=x_1g^{21}+x_2g^{22}+x_3g^{23}+x_4g^{24}=0-x_2+0+0\\
x^3=x_1g^{31}+x_2g^{32}+x_3g^{33}+x_4g^{34}=0+0-x_3+0\\
x^4=x_1g^{41}+x_2g^{42}+x_3g^{43}+x_4g^{44}=0+0+0-x_4\\
\end{align}
A: Remember that in general when we lower the index of a vector, the resulting covector isn't just the original vector turned on its side, this is only the case when the metric is the identity matrix (this is why the distinction between row and column vectors is not always strictly necessary, since in this special case they are just the straight transpose of each  other).
With the metric and vector you have given, $\text{diag}(+,-,-,-)$ and $(x_1,x_2,x_3,x_4)$ you will have:
$$\eta:\begin{pmatrix}x_1\\x_2\\x_3\\x_4 \end{pmatrix}\mapsto (x_1,-x_2,-x_3,-x_4) \tag{1}$$
So when you now contract this with $y^\nu$ you will have picked up the three minus signs you appear to be missing due to the "lowering" operation. Note also that there are two conventions for the Minkowski metric that differ by a sign, you could have also used $\text{diag}(-,+,+,+)$, in which case you would have:
$$\eta:\begin{pmatrix}x_1\\x_2\\x_3\\x_4 \end{pmatrix}\mapsto (-x_1,x_2,x_3,x_4) \tag{2}$$
There is no consensus on which to use, although different fields of physics tend to stick to one convention for various reasons. You do have to make sure however that once you've picked one you stick with it.
