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In the context of four-dimensional spacetime, how does the metric turn a tangent vector into a gradient, and vice versa? By this I mean that I know the metric can be used to raise and lower indices:

$$g^{\mu\nu}V_{\nu}=V^{\mu}$$ and $$g_{\mu\nu}V^{\nu}=V_{\mu}.$$ But my understanding is that, at a point in spacetime, the components of a contravariant vector can be thought of as a tangent vector to a parameterised curve $$V^{\nu}=\frac{dx^{\nu}}{d\lambda}$$ where $\lambda$ is the parameter along the curve. And the components of a covariant vector can be thought of as the gradient of a scalar field $$V_{\nu}=\frac{\partial\phi\left(x\right)}{x^{\nu}}.$$ How does the metric flip $\frac{dx^{\nu}}{d\lambda}$ to $\frac{\partial\phi\left(x\right)}{x^{\nu}}$ or vice versa? To my (non-mathematical) eyes these two things (a tangent to a parameterised curve and a scalar field) look very different to each other.

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  • $\begingroup$ Note that while gradients are a prototypical example of a covector, a general covector can not be expressed as a gradient of a scalar function, i.e. $V_\nu \neq \partial_\nu \phi$ in general. To see this note that taking the derivative of the equation implies the integrability condition $\partial_\mu V_\nu = \partial_\nu V_\mu$ which is definitely not true for a general $V_\mu$. So you can't expect to find a $\phi$ such that $\partial_\mu \phi = g_{\mu\nu}V^\nu$ for some random given $V^\nu$. $\endgroup$ – Michael Brown Sep 9 '13 at 8:58
  • $\begingroup$ If you want to understand things, not just to get used to them, read the maths textbooks. If a vector space V has a non-degenerate bilinear form, then there is a canonical isomorphism between the vector space and its dual, which extends to the corresponding tensor products. Written in coordinates could be confusing if not explained properly. $\endgroup$ – MBN Sep 9 '13 at 12:06
  • $\begingroup$ I've no idea what this means. $\endgroup$ – Peter4075 Sep 9 '13 at 17:02
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The best way to think of this is in terms of maps. A covector is a linear map that turns a vector into a number:

$$ W:\;V^\mu \mapsto W_\mu V^\mu. $$

So if a thing is a linear machine that maps a vector into a scalar we call it a covector. A gradient is an example of such a thing:

$$ \mathrm{d}\phi:\; V^\mu \mapsto \frac{\partial\phi}{\partial x^\mu} V^\mu = \frac{\mathrm{d}\phi}{\mathrm{d}\lambda}, $$

where $V^\mu = \mathrm{d}x^\mu/\mathrm{d}\lambda$. (We can always find a curve tangent to a vector field just by integrating $x^\mu(\lambda) = \int V^\mu(x) \mathrm{d}\lambda$.) But a general covector is not expressible as the gradient of a scalar function: an integrability condition must be satisfied for that.

So let's test $g_{\mu\nu} V^\nu$ to see if it's a covector. We are taking for granted that $g$ is a tensor and $V$ is a vector. Then we contract this into a vector $W^\mu$ and see what happens:

$$ (g_{\mu\nu} V^\nu) W^\mu = g_{\mu\nu} W^\mu V^\nu = W\cdot V, $$

where we recognise the invariant (i.e. scalar) inner product of two vectors. (Recall that $g$ is defined so that this contraction is invariant.) So $g_{\mu\nu} V^\nu$ is a map taking vectors to scalars. You can easily test linearity. Hence $g_{\mu\nu} V^\nu$ is a covector.

If the above mentions integrability condition ($\partial_\sigma (g_{\mu\nu} V^\nu)=\partial_\mu (g_{\sigma\nu} V^\nu)$) is satisfied then you can find a scalar field $\phi$ such that $g_{\mu\nu} V^\nu = \partial_\mu \phi$ by picking some curve $x^\mu(\lambda)$ and integrating $\phi = \int (g_{\mu\nu} V^\nu) (\mathrm{d}x^\mu/d\lambda) \mathrm{d}\lambda$. The integrability condition is necessary because if it doesn't hold then the value of this integral will depend on the full curve and not just its end points, but no curve is better than any other through the same end points so you can't assign a consistent $\phi$.


RE comment:

To get the integrability condition we differentiate $ V_\nu = \partial_\nu \phi $:

$$ \partial_\mu V_\nu = \partial_\mu \partial_\nu \phi = \partial_\nu \partial_\mu \phi = \partial_\nu V_\mu $$

Going the other way: given a covector $W_\mu$ we can make a vector $g^{\mu\nu} W_\nu$ (proof: basically a repeat of the argument given above), then from this vector construct a curve by integrating $$x^\mu(\lambda) = \int g^{\mu\nu} W_\nu \mathrm{d}\lambda$$

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  • $\begingroup$ I'm afraid I don't understand where the “integrability condition” comes from. However, I can live with that as (I think) I get the gist that all gradients are covectors but not all covectors are gradients, and therefore $g_{\mu\nu}V^{\nu}=\partial_{\mu}\phi$ is only sometimes true. How about going the other way? Does $g_{\mu\nu}V^{\nu}$ always give a tangent vector to a parametrised curve, and how would you show that? Apologies if you've already shown this and I've just got lost in the maths. $\endgroup$ – Peter4075 Sep 9 '13 at 10:34
  • $\begingroup$ @Peter4075 See the edits. Hopefully that clears it up. $\endgroup$ – Michael Brown Sep 9 '13 at 11:08

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