# Subtleties in how the metric raises/lowers indices

It is typical for a physicist to write statements like $~g_{\mu\nu}v^{\nu}=v_{\mu}$. But when it is taught that this is just what the metric does (it "raises or lowers indices") with no further insight a student can easily take this for granted. I was taught it like this and now I am trying to find that deeper insight, so please indicate if my thoughts on how the metric raises or lowers indices is correct.

By "the metric lowers indices" what is really meant is that a tangent vector is mapped to a covector. This just means that the result transforms covariantly, as the the change of basis matrix A (as opposed to its inverse $A^{-1}$). To see how this mapping works, consider acting on a vector $V=v^i\partial_i$ with the metric g, where $\{\partial_i\}$ is the coordinate basis with corresponding dual basis $\{dx^i\}$. So we have: $$g=g_{\mu\nu}dx^{\mu}\otimes dx^{\nu}$$ Now act on both sides with the vector V, I will denote a blank argument by - (i.e. "takes something in") $$\implies g(V,-)=g_{\mu\nu}(dx^{\mu}\otimes dx^{\nu})(V,-)$$ $$~~~~~~~~~~~~~~~~=g_{\mu\nu}dx^{\mu}(V) dx^{\nu}(-)$$ $$~~~~~~~~~~~~~~~~~~~=g_{\mu\nu}dx^{\mu}(v^i\partial_i) dx^{\nu}(-)$$ $$~~~~~~~~~~~~~~~~~~~=g_{\mu\nu}v^idx^{\mu}(\partial_i) dx^{\nu}(-)$$ $$~~~~~~~~~~~~=g_{\mu\nu}v^i\delta^{\mu}_{~~~i} dx^{\nu}(-)$$ $$~~~~~~~~=g_{\mu\nu}v^{\mu} dx^{\nu}(-)$$ $$~~:= a_{\nu}dx^{\nu}(-)$$ Where $a_{\nu}:=g_{\mu\nu}v^{\mu}$. It can be easily show that $g_{\mu\nu}v^{\mu}$ transforms covariantly like a (0,1) tensor. So is it the case that when we write $g_{\mu\nu}v^{\mu}=v_{\nu}$, this is bad notation since we shouldn't be using the same letter "$v$" on the LHS and RHS. And then it should be interpreted in the following way; "For every vector $V\in T_pM$, the metric associates a covector (linear functional) by $g:T_pM\rightarrow T^*_pM$ by $V\mapsto g(V,-)$" (Where everything is done at a point $p\in M$). If this is true, we should not consider $g_{\mu\nu}v^{\mu}$ to be the same object as $v_{\nu}$, or to paraphrase mathematically using the above derivation $$V=v^{\nu}\partial_{\nu}\mapsto g(V,-)=g_{\mu\nu}v^{\mu}dx^{\nu}:=a_{\nu}dx^{\nu} \text{ where } a_{\nu}\neq v_{\nu}.$$

• Where is the question here? If you're asking if your reasoning is right, yes, it is. But metric raising and lowering is so common we might as well call the vector $v$ and one-form $v$ by the same letter. – knzhou Apr 27 '17 at 4:16
• OK. I understand your point. What would you call the object $v_\nu$? – Prahar Apr 27 '17 at 5:08
• My question was if my reasoning was right, because we were only told (almost like a definition) that $g_{\mu\nu}v^{\mu}=v_{\nu}$, without actually showing what is going on and how it happens. I agree 100% that after it is understood that this is whats going on behind the scenes then we can drop all intermediate steps and just write $g_{\mu\nu}v^{\mu}=v_{\nu}$, but only after it is understood. I also agree that we may aswell call it $v_{\nu}$, but again after it is understood. – NormalsNotFar Apr 27 '17 at 5:23
• If you write it out without the Einstein notation, you will see from your indices where the raising and lowering comes from. In my first course in GR we did bottom-up development. Today it is more likely to be from a more abstract approach to the math. Much like Taylor's book on mechanics versus a course from Suskind. – C. Towne Springer Apr 27 '17 at 5:31
• You could simply define $v_\mu \equiv g_{\mu\nu} v^\nu$. If you want to define it in some other other way, tell me what that is first. – Prahar Apr 27 '17 at 5:32

Actually it's the opposite. $$v_\mu$$ and $$v^\mu$$ are components of two completely different objects, a covector and a vector, so their labeling should not fool you: they are in principle two different things, and unless someone defines some rule to connect them, they are two entities completely unrelated from each other.
Every environment brings its own rule, and many books/authors adopt the convention that the dual vector to some vector $$v$$, chosen by the isomorphism, shall have the same label of the original vector:
• this happens in quantum mechanics where the ket $$|\psi\rangle$$ goes in the bra $$\langle \psi|$$, via the inner product on Hilbert space,
• this happens in general relativity, where [a vector $$v=v^\mu e_\mu$$] goes in [a 1-form $$v=v_\mu e^\mu$$] thanks to the inner product defined by the metric);
In this way, we have selected a way, which finds its roots in the basic gymnastics of the theory we are developing (i.e. the modulus square of a vector, or of a wavefunction), and in this way we connected univocally the otherwise totally distinct entities $$v_\mu$$ and $$v^\mu$$.