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Let $\Omega^*$ be the algebra generated by $dx_1,\dots,dx_n$ subject to the relations $(dx_i)^2=0$ and antisymmetry.

Does this form the cotangent space $T^*$?

I thought it might, since the tangent space $T$ has basis $\left\{\frac{\partial}{\partial x_1},\dots,\frac{\partial}{\partial x_n}\right\}$

But I wasn't sure how to see that $dx_1\frac{\partial}{\partial x_1}=1$ (assuming this behaves like a traditional dual space.)

Context: Bott and Tu says that the exterior differentiation $d$ gives us $d(0-\!\text{forms})=\text{gradient}$, and normally we write (for $3$-dim space say) $$Df=\left(\frac{\partial f}{\partial x},\dots,\frac{\partial f}{\partial z}\right)$$

Where here the exterior differentiation gives us: $$df=\frac{\partial f}{\partial x} dx+ \dots + \frac{\partial f}{\partial z} dz$$

Which in the basis $\{dx,dy,dz\}$ is the same, where I thought that this was the basis of a cotangent space perhaps.

Earth Cracks
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  • The relation $dx_i^2 = 0$ is a consequence of antisymmetry. Also, are you asking about the (co)tangent space to a smooth manifold at some fixed point? – Travis Willse Jun 18 '17 at 11:00
  • @Travis I suppose so, but I'm so unsure that I wasn't able to specify further. I had done a physics course that weakly touched these things, and now I am looking at Bott and Tu, trying to fix them together – Earth Cracks Jun 18 '17 at 11:01
  • As for showing that $dx_1\left(\frac{\partial}{\partial x_1}\right) = 1$, this depends on your definitions of those objects. – Travis Willse Jun 18 '17 at 11:01

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