2) What you did for 2.1 is correct.

I’m not sure what you mean in 2.2 by “cancel.” Remember that $dx(\frac{\partial}{\partial x}) = 1$ and $dx(\frac{\partial}{\partial y}) = 0$, etc.

For 2.3, your perspective from (a) is the correct one. At each point, you can evaluate what $V$ and $\alpha$ are, then you can apply $\alpha$ to $V$.

]]>\[

(\alpha \wedge \beta)(X,Y)_p := \alpha_p(X_p) \beta_p(Y_p) – \alpha_p(Y_p) \beta_p(X_p),

\]

i.e., the value of the resulting function at the point \(p\) is obtained by grabbing the 1-form and vector values at \(p\) and evaluating the usual expression. (You may start to see why we generally omit the subscript “\(p\)”!)

]]>1) The application of a differential 1-form $\alpha$ to a Vector field U results in a scalar function $R^n \rightarrow R$ that measures the non-normalized projection (i.e. dot product) of each vector of $U_{x, y}$ in relation to the vector $\alpha(x), \alpha(v)$. Therefore, $\alpha$ will result in different values for the angles (also influenced by magnitude) between $U_{x, y}$ and $\alpha(x), alpha(v)$:

– negative for angle $>\frac{1}{2}\pi$

– zero for angle $ = 0$

– positive for inner angle $<\frac{1}{2}\pi$

2) I am confused with slide 46 and ex. 2.2. I will use an example. Let's say I have a vector field $V = y \frac{\partial}{\partial x} – x\frac{\partial}{\partial y}$ and a differential 1-form $\alpha = x dx + 2y dy$.

-2.1: If I evaluate U at (3, 5), I get the vector $V_{3, 5} = 5\frac{\partial u}{\partial x} – 3\frac{\partial u}{\partial y}$

-2.2 Can I evaluate $\alpha(.)$ for any input with coordinates or only for vector-field? If I apply $\alpha$ for regular coordinates: $\alpha_{3, 5} = 3 dx + 10 dy$. By definition, I should get a real number, but in this case, I get kind of a vector. I feel I can only apply $\alpha$ to vector fields, because dx will cancel $\frac{\partial}{\partial x}$ and dy will cancel $\frac{\partial}{\partial y}$, resulting in the real number. Is this interpretation correct?

-2.3 If I evaluate $\alpha(V)$, I foresee two options… (a) I treat (x, y) as the original coordinates (x, y) in $R^3$, and $\alpha(V) = xy – 2yx = -xy$; (b) I treat (x, y) as the coordinates of the argument, so I have $\alpha(V) = y – 2x$. I am tending to the interpretation (a).