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3

Yes. In fact, this is true for any $k$-form, regardless of whether or not it is in the image of $d$. To see this, let $\omega$ be any smooth $k$-form on $N$. Let $U\subseteq N$ denote the open set consisting of all $p\in N$ for which $\omega_p$ is not identically $0$. We will assume $U\neq \emptyset$, and then prove that $f^\ast \omega$ is not the $0$ ...


3

This seems to be false, even if one assumes that $\Sigma$ is isometrically embedded. Counterexample: Let $M$ be the $2$-sphere and $\Sigma$ some great circle in $M$; the volume form $\Omega$ of $\Sigma$ is some parallel one-form on $\Sigma$. $\Omega$ can't possibly extend even locally to some neighborhood of $M$, because $M$ doesn't admit even locally ...


3

It's just componentwise. Fix a basis $e_i$ with dual basis $\theta^i$ and define $$\int \omega = \sum_i \left(\int \theta^i \circ \omega\right) e_i.$$ If you have another basis $v_i$ with dual $\pi^i$, then let $A$ be the change of basis matrix such that $v_j = \sum_i A_{ji} e_i$. A little bit of linear algebra tells us that the dual bases are then related ...


3

Interesting. That exercise got me thinking for awhile, I had no choice but to open everything in coordinates, but I nailed it. I wonder if there's a better way to do it. What you did so far is correct. Since we want to compare these forms in ${\bf x}(D)$ (and you should really use ${\bf x}$ instead of $x$ here, for reasons that'll be clear shortly), it is ...


2

Suppose the charts on $Y$ follow coordinates $w \in \mathbb{C}^n$. Then we may write $\omega$ locally in a chart as $$ \omega(w) = \frac{\partial^2 f}{\partial w \partial \bar w }(w) dw \wedge d\bar w $$ Now if we have a nice holo map $F: X \to Y$ (i.e. structure preserving), the pull back is just as you've written, $F^* \omega = \omega \circ F$. This ...


2

Following the hint, if $F: \partial W \to Y$ is a smooth map that extends to all $W$ and $\omega$ is a closed $k$-form on $Y$, then $d\omega=0$, thus Stokes Theorem gives $$ \int_{\partial W} F^*\omega = \int_{W} d(F^*\omega) =\int_{W} F^*(d\omega) = \int_W 0 = 0 \tag{1} $$ Now let $F: X \times I \to Y$ be an homotopy between $f_0$ and $f$. Since ...


2

First, we should define tensors. There are various types of tensors, and we focus on those which are relevant to your question. A tensor of type $(k,0)$ is a map$$T:\Gamma(TM)\times\ldots\times\Gamma(TM)\to C^\infty(M),$$which is $C^\infty(M)$-linear in every coordinate. In other words, $T$ eats $k$ vector fields and spits out a function. The linearity means ...


2

Let's recall the definition of the module of differentials on a $\mathbb{C}$-algebra $R$: take the free $R$-module on the symbols $\{ dr \colon r \in R \}$ and mod out by the relations $d(r+s ) = dr + ds$, $d(rs) = sdr + r ds$, and $da = 0$ for $a \in \mathbb{C}$. Then, the sheaf $\Omega_{R/\mathbb{C}}^1$ on $\textrm{Spec}(R)$ is the "tilde" of this module. ...


2

Why it has to do with index lowering: Because it maps the vector of components $p^i$ to the form of components $p_i$. Why it has to do with the Euclidean metric: One cannot just copy the components of a vector and write them into a differential form. They are different objects in different spaces. What really happens is that you get a linear map ...


2

In that sentence, I was referring to the tangent-cotangent isomorphism, defined on pages 341-343 of the book. It probably would have been a good idea to include a page reference.


2

That's correct! We say a form $\omega$ is closed if $d\omega = 0$, and we say that $\omega$ is exact if $\omega = d\eta$ for some form $\eta$. Your remark says, in this terminology, that every exact form is closed. However, the converse is not true: not every closed form is exact. (Here, I am referring to forms defined on our whole manifold - the Poincare ...


2

Notice that for $\Omega_i$ it is always possible to write: $$ \Omega_{i} = \sum_{j\neq i} a_j \Omega_j $$ for some constants $a_j$. For example: $\Omega_1 = \Omega_2 + \Omega_4 -\Omega_3 $. So if all $j\neq i$ forms are exact, so will be $\Omega_i$.


1

Let $V$ denote an $n$-dim vector space (over $\mathbb{R}$, say). Recall that a $p$-form $\alpha \in \Lambda^p(V^*)$ is decomposable (or simple) iff it can be written as a wedge product of $1$-forms -- i.e., $\alpha = \omega_1 \wedge \cdots \wedge \omega_p$, with each $\omega_i \in \Lambda^1(V^*) = V^*$. Definition: Let $\alpha \in \Lambda^p(V^*)$ be a ...


1

You certainly don't need stereographic coordinates. By definition, you need to check that $\omega$ is a smooth 2-form, $d \omega=0$, and $\omega$ is non-degenerate, i.e. for all $v \in T_u S^2$, there exists $w \in T_u S^2$ such that $\omega_u(v,w)\neq 0$. You could try to verify (1) using stereographic coordinates, but that seems unnecessary. The ...


1

By Sard's theorem, there is a dense subset $U \in N$ so that $Df$ is sujective on $f^{-1}(U)$. Then $d\omega$ is zero on $U$: this is a general fact in linear algebra: Let $L : V\to W$ be a surjective linear map. If $$L^*\alpha(X, Y, Z) = \alpha(LX, LY, LZ) = 0$$ for all $X, Y, Z\in V$, then $$\alpha (a, b, c) = 0$$ for all $a, b, c \in W$, as $L$ is ...


1

There is. Let $\pi : \mathbb R^4\to \mathbb R^2$ be the projection to the $y_1, y_2$-plane and $i : \mathbb R^2 \to \mathbb R^4$ be $(y_1, y_2) \mapsto (0,0, y_1, y_2)$. Then your $P_{[dy_1,dy_2]}$ is $\pi^* \circ i^* = (i\circ\pi)^*$.


1

I guess what I write here is wrong. But, I'm not sure exactly why, so I post it here in the hope someone points out the error of my ways. The two-form $\omega = e^{xy}dx \wedge dy$ may be written as $\omega = e^{r^2 \sin \theta \cos \theta} r \, dr \wedge d\theta$. Then, in polar coordinates, the unit circle is $r=1$ and $\partial / \partial r$ is an ...


1

That's precisely correct! See This wikipedia page on closed and exact forms for more details.


1

Metric can be thought of as a mapping that takes two vector fields as arguments and produces scalar field. By fixing one vector field (call it $X$) you get a map from vector fields to scalars, that is a diferential form $gX$. This differential form will map vector field $Y$ to its scalar product with $X$. If you use Cartesian coordinates in Euclidean space ...


1

Note that, if $$A(x,y)=\frac{f(x,y)}{xf(x,y)+yg(x,y)}$$ and $$B(x,y)=\frac{f(x,y)}{xf(x,y)+yg(x,y)},$$ you can derive and find that $$A_y=\frac{yf_yg-fg-yfg_y}{(xf+yg)^2}$$ and $$B_x=\frac{xfg_x-fg-xf_xg}{(xf+yg)^2}.$$ Theorem: If $\omega=Adx+Bdy$ a 1-form is $C^1$ such that $A_y=B_x$, then $\omega$ is closed. Hint: Use that $f$ and $g$ are ...



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