# Definition of exterior derivative

In Sean Carroll's GR book, pg 84, the exterior derivative $$d$$ is defined as $$(dA)_{\mu_1\mu_2...\mu_{p+1}} = (p+1) \partial_{[\mu_1}A_{\mu_2...\mu_{p+1}]}\tag{2.76}$$ where $$A$$ is a $$p$$-form and the RHS is the appropiately normalized and antisymmetrized partial derivative.

I know that to antisymmetrize a tensor, we mean something like $$t_{[ab]}=\frac{1}{2}(t_{ab}-t_{ba}).$$

But the partial derivative of a tensor is generally not a tensor. So what does it mean to antisymmetrize the partial derivative of a tensor? In other words, what does $$\partial_{[\mu_1}A_{\mu_2...\mu_{p+1}]}$$ mean?

• – Qmechanic Mar 7 at 11:20

## 1 Answer

When people say the partial derivative of a tensor is not a tensor, what they mean is that when you change your coordinate system the transformation is not just a simple tensor transformation, e.g. $$t_{a'b'}=\frac{\partial{x^a}}{\partial x^{a'}}\frac{\partial{x^b}}{\partial x^{b'}}t_{ab}$$ A partial derivative of a tensor does not typically satisfy this transformation rule, but that's completely okay. Anti-symmetrizing an object with indices has nothing to do with transformations. It is defined in the same way as you state in your question.

Note however that the exterior derivative does in fact satisfy the tensor transformation law (you can easily show this as an exercise). The point though is that the definition of anti-symmetrizion is just an operation on indices that is well defined regardless of transformation properties.