# 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

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.