I only understood this after taking a class in general relativity (GR), differential geometry and quantum field theory (QFT). The essence is quite trivial, actually:

You have a theory that is invariant under some symmetry group. So in quantum electrodynamics you have a Lagrangian density for the fermions (no photons yet) $$ \mathcal L = \bar\psi(x) [\mathrm i \gamma^\mu \partial_\mu - m] \psi(x) \,.$$ This $\bar\psi $ is just $\psi^\dagger \gamma^0$, important is that it is complex conjugated. The fact that it is a four-vector in spin-space is of no concern here. What one can do now is transform $\psi \to \exp(\mathrm i \alpha) \psi$. Then $\bar\psi \to \bar\psi \exp(-\mathrm i \alpha)$ and the Lagrangian will be invariant. There you have a global symmetry.

Now promote the symmetry to a local one, why not? Instead of a global $\alpha$ one now has $\alpha(x)$. The problem is that when we transform now, one picks up the $\partial_\mu \alpha(x)$ with the chain and product rules of differentiation. That seems like a technical complication at first.

There is a more telling way to see this:
You take a deriviative of a field $\psi(x)$. This means taking a difference quotient like $$ \partial_\mu \psi(x) = \lim_{\epsilon \to 0} \frac{\psi(x + \epsilon \vec e_\mu) - \psi(x)}{\epsilon} \,.$$ This works just fine with a global transformation. But with the local transformation, you basically subtract two values that are gauged differently. In differential geometry you have that the tangent spaces at the different points of the manifold are different and therefore one cannot just compare vectors by their components. One needs a connection with connection coefficients to provide parallel transport. It is similar here. We now have promoted $\phi$ from living on $\mathbb R^4$ to living in the bundle $\mathbb R^4 \times S^1$ as we have an U(1) gauge group. Therefore we need some sort of connection in order to transport the transformed $\phi$ from $x + \epsilon \vec e_\mu$ to $x$. This is where one has to introduce some connection which is $$ \partial_\mu \to \mathrm D_\mu := \partial_\mu + \mathrm i A_\mu \,.$$

If you plug that into the Lagrange density to make it $$ \mathcal L = \bar\psi(x) [\mathrm i \gamma^\mu \mathrm D_\mu - m] \psi(x)$$ and then choose $A_\mu = \partial_\mu \alpha$ you will see that the Lagrangian density does stay invariant even under local transformations as the connection coefficient will just subtract the unwanted term from the product/chain rule.

In general relativity you have the symmetry under arbitrary diffeomorphism, the price is that you have to change the derivative to a connection, $$ \partial \to \nabla := \partial + \Gamma + \cdots \,.$$

I only understood this after taking a class in general relativity (GR), differential geometry and quantum field theory (QFT). The essence is just a change of coordinates systems that needs to be reflected in the derivative. I'll explain what I mean.

You have a theory that is invariant under some symmetry group. So in quantum electrodynamics you have a Lagrangian density for the fermions (no photons yet) $$ \mathcal L = \bar\psi(x) [\mathrm i \gamma^\mu \partial_\mu - m] \psi(x) \,.$$ This $\bar\psi $ is just $\psi^\dagger \gamma^0$, important is that it is complex conjugated. The fact that it is a four-vector in spin-space is of no concern here. What one can do now is transform $\psi \to \exp(\mathrm i \alpha) \psi$ with some $\alpha \in \mathbb R$. Then $\bar\psi \to \bar\psi \exp(-\mathrm i \alpha)$ and the Lagrangian will be invariant as the derivative does not act on the exponential function, it is just a phase factor. There you have a global symmetry.

Now promote the symmetry to a local one, why not? Instead of a global $\alpha$ one now has $\alpha(x)$. This means we choose a different $\alpha$ at each point in spacetime. The problem is that when we transform now, one picks up the $\partial_\mu \alpha(x)$ with the chain and product rules of differentiation. That seems like a technical complication at first.

There is a more telling way to see this:
You take a deriviative of a field $\psi(x)$. This means taking a difference quotient like $$ \partial_\mu \psi(x) = \lim_{\epsilon \to 0} \frac{\psi(x + \epsilon \vec e_\mu) - \psi(x)}{\epsilon} \,.$$ This works just fine with a global transformation. But with the local transformation, you basically subtract two values that are gauged differently. In differential geometry you have that the tangent spaces at the different points of the manifold are different and therefore one cannot just compare vectors by their components. One needs a connection with connection coefficients to provide parallel transport. It is similar here. We now have promoted $\phi$ from living on $\mathbb R^4$ to living in the bundle $\mathbb R^4 \times S^1$ as we have an U(1) gauge group. Therefore we need some sort of connection in order to transport the transformed $\phi$ from $x + \epsilon \vec e_\mu$ to $x$. This is where one has to introduce some connection which is $$ \partial_\mu \to \mathrm D_\mu := \partial_\mu + \mathrm i A_\mu \,.$$

If you plug that into the Lagrange density to make it $$ \mathcal L = \bar\psi(x) [\mathrm i \gamma^\mu \mathrm D_\mu - m] \psi(x)$$ and then choose $A_\mu = \partial_\mu \alpha$ you will see that the Lagrangian density does stay invariant even under local transformations as the connection coefficient will just subtract the unwanted term from the product/chain rule.

In general relativity you have the symmetry under arbitrary diffeomorphism, the price is that you have to change the derivative to a connection, $$ \partial \to \nabla := \partial + \Gamma + \cdots \,.$$

I only understood this after taking a class in general relativity (GR), differential geometry and quantum field theory (QFT). The essence is quite trivial, actually:

You have a theory that is invariant under some symmetry group. So in quantum electrodynamics you have a Lagrangian density for the fermions (no photons yet) $$ \mathcal L = \bar\psi(x) [\mathrm i \gamma^\mu \partial_\mu - m] \psi(x) \,.$$ This $\bar\psi $ is just $\psi^\dagger \gamma^0$, important is that it is complex conjugated. The fact that it is a four-vector in spin-space is of no concern here. What one can do now is transform $\psi \to \exp(\mathrm i \alpha) \psi$. Then $\bar\psi \to \bar\psi \exp(-\mathrm i \alpha)$ and the Lagrangian will be invariant. There you have a global symmetry.

Now promote the symmetry to a local one, why not? Instead of a global $\alpha$ one now has $\alpha(x)$. The problem is that when we transform now, one picks up the $\partial_\mu \alpha(x)$ with the chain and product rules of differentiation. That seems like a technical complication at first.

There is a more telling way to see this:
You take a deriviative of a field $\psi(x)$. This means taking a difference quotient like $$ \partial_\mu \psi(x) = \lim_{\epsilon \to 0} \frac{\psi(x + \epsilon \vec e_\mu) - \psi(x)}{\epsilon} \,.$$ This works just fine with a global transformation. But with the local transformation, you basically subtract two values that are gauged differently. In differential geometry you have that the tangent spaces at the different points of the manifold are different and therefore one cannot just compare vectors by their components. One needs a connection with connection coefficients to provide parallel transport. It is similar here. We now have promoted $\phi$ from living on $\mathbb R^4$ to living in the bundle $\mathbb R^4 \times S^1$ as we have an U(1) gauge group. Therefore we need some sort of connection in order to transport the transformed $\phi$ from $x + \epsilon \vec e_\mu$ to $x$. This is where one has to introduce some connection which is $$ \partial_\mu \to \mathrm D_\mu := \partial_\mu + \mathrm i A_\mu \,.$$

If you plug that into the Lagrange density to make it $$ \mathcal L = \bar\psi(x) [\mathrm i \gamma^\mu \mathrm D_\mu - m] \psi(x)$$ and then choose $A_\mu = \partial_\mu \alpha$ you will see that the Lagrangian density does stay invariant even under local transformations as the connection coefficient will just subtract the unwanted term from the product/chain rule.

In general relativity you have the symmetry under arbitrary diffeomorphism, the price is that you have to change the derivative to a connection, $$ \partial \to \nabla + \Gamma + \cdots \,.$$

I only understood this after taking a class in general relativity (GR), differential geometry and quantum field theory (QFT). The essence is quite trivial, actually:

You have a theory that is invariant under some symmetry group. So in quantum electrodynamics you have a Lagrangian density for the fermions (no photons yet) $$ \mathcal L = \bar\psi(x) [\mathrm i \gamma^\mu \partial_\mu - m] \psi(x) \,.$$ This $\bar\psi $ is just $\psi^\dagger \gamma^0$, important is that it is complex conjugated. The fact that it is a four-vector in spin-space is of no concern here. What one can do now is transform $\psi \to \exp(\mathrm i \alpha) \psi$. Then $\bar\psi \to \bar\psi \exp(-\mathrm i \alpha)$ and the Lagrangian will be invariant. There you have a global symmetry.

Now promote the symmetry to a local one, why not? Instead of a global $\alpha$ one now has $\alpha(x)$. The problem is that when we transform now, one picks up the $\partial_\mu \alpha(x)$ with the chain and product rules of differentiation. That seems like a technical complication at first.

There is a more telling way to see this:
You take a deriviative of a field $\psi(x)$. This means taking a difference quotient like $$ \partial_\mu \psi(x) = \lim_{\epsilon \to 0} \frac{\psi(x + \epsilon \vec e_\mu) - \psi(x)}{\epsilon} \,.$$ This works just fine with a global transformation. But with the local transformation, you basically subtract two values that are gauged differently. In differential geometry you have that the tangent spaces at the different points of the manifold are different and therefore one cannot just compare vectors by their components. One needs a connection with connection coefficients to provide parallel transport. It is similar here. We now have promoted $\phi$ from living on $\mathbb R^4$ to living in the bundle $\mathbb R^4 \times S^1$ as we have an U(1) gauge group. Therefore we need some sort of connection in order to transport the transformed $\phi$ from $x + \epsilon \vec e_\mu$ to $x$. This is where one has to introduce some connection which is $$ \partial_\mu \to \mathrm D_\mu := \partial_\mu + \mathrm i A_\mu \,.$$

If you plug that into the Lagrange density to make it $$ \mathcal L = \bar\psi(x) [\mathrm i \gamma^\mu \mathrm D_\mu - m] \psi(x)$$ and then choose $A_\mu = \partial_\mu \alpha$ you will see that the Lagrangian density does stay invariant even under local transformations as the connection coefficient will just subtract the unwanted term from the product/chain rule.

In general relativity you have the symmetry under arbitrary diffeomorphism, the price is that you have to change the derivative to a connection, $$ \partial \to \nabla := \partial + \Gamma + \cdots \,.$$