[[Noteworthy Uses]]:: *[[Noteworthy Uses]]* [[Proved By]]:: *[[Proved By|Crucial Dependencies]]* Intuition:: [[dual transform of forms#^d39630|Inspiration is here]] Specializations:: *[[Specializations]]* Generalizations:: *[[Generalizations]]* ---- - Let $\v x := (x_{1},\dots,x_{k})$ denote a generic point in $\rr ^{k}$ and let $A \subset \rrk$ be [[open set|open in]] $\rrk$; - Let $\v y = (y_{1},\dots,y_{k})$ denote a generic point in $\rrn$ and let $B \subset \rrn$ be [[open set|open in]] $\rrn$; then $dx_{i}$ and $dy_{i}$ denote the elementary $1$-forms on $\rrk$ and $\rrn$, respectively; - Let $\alpha: A \subset \rrk \to B \subset \rrn$ be a [[smooth]] map. > [!theorem] Theorem. ([[computation of dual transform of forms]]) > Then, >1. ($1-\text{form}$) $\alpha ^{*}(dy_{i})=d\alpha_{i} = \sum_{j=1}^{k}D_{j} \alpha_{i} \ dx_{j}$ ; >2. ($k-\text{form}$) For $I = (i_{1},\dots,i_{k}) \in$ [[set of all ascending k-tuples from 1 to n]], $\alpha ^{*} (dy_{I}) = \left( \det \frac{ \partial \alpha_{I} }{ \partial \v x } \right)\ dx_{I}$, where $dy_{I}=dy_{i_{1}}\wedge \dots \wedge dy_{i_{k}} \and \frac{ \partial \alpha_{I} }{ \partial \v x } = \frac{ \partial (\alpha_{i_{1}} , \dots, \alpha_{i_{k}})}{ \partial (x_{1},\dots,x_{k} )} .$ >3. ($\ell-\text{form}$) For $I=(i_{1},\dots,i_{\ell}) \in \asc_{\ell ,n}$, $\alpha ^{*}(dy_{I})=\sum_{J \in \asc_{\ell, k}}^{}\left( \det \frac{ \partial \alpha_{I} }{ \partial x_{J} } \right) \ dx_{J},$ >where $J=(j_{1},\dots,j_{\ell})$ and $\frac{ \partial \alpha_{I} }{ \partial x_{J} }=\frac{ \partial (\alpha_{i_{1}},\dots, \alpha_{i_{\ell}}) }{ \partial (x_{j_{1}},\dots, x_{j_{\ell}}) }$ . > [!proof]- Proof. ([[computation of dual transform of forms]]) > 1. Set $\v y = \alpha(\v x)$. Let $\v v \in \rrk$ be arbitrary. Then compute $\begin{align} > \alpha ^{*}(dy_{i})(\v x)(\v v)= & \big(dy_{i} (\alpha(\v x))\big) \ (D\alpha(\v x )\v v) \\ > = & (i^{th} \text{ component of the vector } D\alpha(\v x) \v v )\\ > = & \sum_{j=1}^{k} D_{j}\alpha_{i}(\v x) v_{j} \\ > \end{align}$ > and since $v_{j}=dx_{j}(\v v)$, this implies $\alpha ^{*}(dy_{i})=\sum_{j=1}^{k}(D_{j}\alpha _{i})dx_{j}$ > which by definition equals $d\alpha_{i}$ as desired. > > 2. Since the [[dual transform of forms#^f2843f|dual transform of forms respects the]] [[algebra of alternating multilinear forms|wedge product]] we have $\begin{align} > \alpha ^{*}(dy_{i_{1}} \wedge \dots \wedge dy_{i_{k}}) = & \alpha ^{*}(dy_{i_{1}}) \wedge \dots \wedge \alpha ^{*}(dy_{i_{k}}) \\ > = & d\alpha_{i_{1}} \wedge \dots \wedge d\alpha_{i_{k}} \text{ (part 1)} \\ > = & \det \begin{bmatrix} > — & D\alpha_{i_{1}} & — \\ & \vdots & \\ — & D\alpha_{i_{k}} & — > \end{bmatrix} \ dx_{1} \wedge \dots \wedge dx_{k}, > \end{align}$ > where in the last step we used the **[[396 HW4 Q1 Lemma]]**. Since $\begin{bmatrix} > — & D\alpha_{i_{1}} & — \\ & \vdots & \\ — & D\alpha_{i_{k}} & — > \end{bmatrix} = \begin{bmatrix} > \frac{ \partial \alpha_{i_{1}} }{ \partial x_{1} } & \dots & \frac{ \partial \alpha _{i_{1}} }{ \partial x_{k} } \\ \vdots& \ddots & \vdots \\ > \frac{ \partial \alpha_{i_{k}} }{ \partial x_{1}} & \dots & \frac{ \partial \alpha_{i_{k}} }{ \partial x_{k} }\end{bmatrix} := \frac{ \partial \alpha_{I} }{ \partial \v x }, $ > this is the result. ---- #### ----- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```