Updates

Signed-off-by: Marcello Seri <marcello.seri@gmail.com>

Author: mseri

5-tensors.tex

@@ -70,7 +70,7 @@ \section{Tensors}
   Let $V$ an $n$-dimensional vector space, $\tau_1\in T_s^r(V)$, $\tau_2\in T_{s'}^{r'}(V)$.
   We define the \emph{tensor product} $\tau_1\otimes\tau_2$ as the $(r+r', s+s')$-tensor defined by
   \begin{align}
-     & \tau_1\otimes\tau_2(\omega^1,\ldots,\omega^{r+r'}, v_1,\ldots,v_{s+s'})                                                    \\
+     & \tau_1\otimes\tau_2(\omega^1,\ldots,\omega^{r+r'}, v_1,\ldots,v_{s+s'})                                                          \\
      & = \tau_1(\omega^1,\ldots,\omega^{r}, v_1,\ldots,v_{s}) \cdot \tau_2(\omega^{r+1},\ldots,\omega^{r+r'}, v_{s+1},\ldots,v_{s+s'}).
   \end{align}
 \end{definition}
@@ -252,7 +252,7 @@ \section{Tensors}
     \cI = \cI_g : T_s^r (V) \to T_r^s(V) \\
     \cI : \tau \mapsto \tau \circ (\LaTeXunderbrace{{\cdot}^\flat, \ldots, {\cdot}^\flat}_{r\mbox{ times}}, \LaTeXoverbrace{{\cdot}^\sharp, \ldots, {\cdot}^\sharp}^{s\mbox{ times}}).
   \end{align}
-    In general, one can use the metric tensor to raise or lower arbitrary indices, changing the tensor type from $(r,s)$ to $(r+1, s-1)$ or $(r-1, s+1)$.
+  In general, one can use the metric tensor to raise or lower arbitrary indices, changing the tensor type from $(r,s)$ to $(r+1, s-1)$ or $(r-1, s+1)$.
 
   A neat application of this is showing that a non-degenerate bilinear map $g\in T_2^0(V)$ can be lifted to a non-degenerate bilinear map on arbitrary tensors, that is
   \begin{equation}
@@ -270,13 +270,13 @@ \section{Tensors}
     \item Fix a basis for $V$. What does $I_g$ in the previous remark look like with respect to this basis?
     \item Write down $G$ with respect to the basis from the previous point.
   \end{enumerate}
-      %In coordinates the one mapping $(r,s)$-tensors to $(r+s, 0)$-tensors is $I_g(\tau) = \tau_{i_1\ldots i_r}^{j_1\ldots j_s} g_{j_1 j_{r+1}} \ldots g_{j_s j_{r+s}}}$
-      %If \((g_{ij})\) and \((g^{ij})\) are the matrix element of the matrices representing metric tensor and its inverse resp, then \[ G(\sigma, \tau) = \langle\sigma, \tau\rangle_g := g^{k_1 l_1}\cdots g^{k_rl_r} g_{i_1j_1} \cdots g_{i_sj_s} \sigma_{k_1,\ldots,k_r}^{i_1,\ldots,i_s} \tau_{l_1,\ldots,l_r}^{j_1,\ldots,j_s} \]
+  %In coordinates the one mapping $(r,s)$-tensors to $(r+s, 0)$-tensors is $I_g(\tau) = \tau_{i_1\ldots i_r}^{j_1\ldots j_s} g_{j_1 j_{r+1}} \ldots g_{j_s j_{r+s}}}$
+  %If \((g_{ij})\) and \((g^{ij})\) are the matrix element of the matrices representing metric tensor and its inverse resp, then \[ G(\sigma, \tau) = \langle\sigma, \tau\rangle_g := g^{k_1 l_1}\cdots g^{k_rl_r} g_{i_1j_1} \cdots g_{i_sj_s} \sigma_{k_1,\ldots,k_r}^{i_1,\ldots,i_s} \tau_{l_1,\ldots,l_r}^{j_1,\ldots,j_s} \]
 \end{exercise}
 
-\begin{remark}
+\begin{remark}\label{rem:gradedtensoralgebra}
   Interestingly, even though none of the tensor spaces $T_s^r(V)$ are algebras, the map $\otimes$ makes the collection of all tensor spaces
-  \marginnote{This is a so-called \emph{graded algebra} since $\otimes :  T_s^r(V)\times T_{s'}^{r'}(V) \to T_{s+s'}^{r+r'}(V)$ in some sense moves along the structure of the indices.}
+  \marginnote{This is a so-called \emph{graded algebra} since $\otimes :  T_s^r(V)\times T_{s'}^{r'}(V) \to T_{s+s'}^{r+r'}(V)$ in some sense moves along the structure of the indices. Technically, an algebra is graded if its additive group can be decomposed into a direct sum of subgroups. Any element has a degree defined by the subgroup it belongs to.}
   \begin{equation}
     T(V) := \bigoplus_{r,s\geq 0} T_s^r(V), \qquad T_0^0(V):= \R,
   \end{equation}
@@ -290,12 +290,12 @@ \section{Tensors}
   Let $V$ be a vector space and fix $r,s\geq0$.
   For $h\leq r$ and $k\leq s$, we define the \emph{$(h,k)$-contraction} of a tensor as the linear mapping $T_s^r(V)\to T_{s-1}^{r-1}(V)$ defined through
   \begin{align}
-    v_1 & \otimes\cdots\otimes v_r\otimes\omega^1\otimes\cdots\otimes\omega^s \\
+    v_1 & \otimes\cdots\otimes v_r\otimes\omega^1\otimes\cdots\otimes\omega^s                                                                                                             \\
         & \mapsto \omega^k(v_h)\, v_1\otimes\cdots\otimes v_{h-1}\otimes v_{h+1}\cdots\otimes v_r\otimes\omega^1\otimes\cdots\otimes\omega^{k-1}\otimes\omega^{k+1}\cdots\otimes\omega^s,
   \end{align}
   and then extended by linearity, thus mapping $\tau \mapsto \widetilde\tau$ where
   \begin{align}
-    \widetilde\tau & (\nu^1,\ldots,\nu^{r-1}, v_1,\ldots,v_{s-1}) \\
+    \widetilde\tau & (\nu^1,\ldots,\nu^{r-1}, v_1,\ldots,v_{s-1})                                                                                                       \\
                    & = \tau(\nu^1,\ldots,\LaTeXunderbrace{e^i}_{h\mbox{th index}},\ldots,\nu^{r-1},w_1,\ldots,\LaTeXunderbrace{e_i}_{k\mbox{th index}},\ldots,w_{s-1}).
   \end{align}
 \end{definition}
@@ -304,9 +304,9 @@ \section{Tensors}
   It is common to use an hat to denote elements that have been removed from the tensor product.
   For instance, the contraction above would look like
   \begin{align}
-    v_1 & \otimes\cdots\otimes v_r\otimes\omega^1\otimes\cdots\otimes\omega^s \\
-        & \mapsto \omega^k(v_h)\, v_1\otimes\cdots\otimes v_{h-1}\otimes v_{h+1}\cdots\otimes v_r\otimes\omega^1\otimes\cdots\otimes\omega^{k-1}\otimes\omega^{k+1}\cdots\otimes\omega^s \\
-        &\qquad  =: \omega^k(v_h)\, v_1\otimes\cdots\otimes \widehat{v}_{h} \otimes \cdots\otimes v_r\otimes\omega^1\otimes\cdots\otimes\widehat{\omega}^{k}\otimes\cdots\otimes\omega^s.
+    v_1 & \otimes\cdots\otimes v_r\otimes\omega^1\otimes\cdots\otimes\omega^s                                                                                                              \\
+        & \mapsto \omega^k(v_h)\, v_1\otimes\cdots\otimes v_{h-1}\otimes v_{h+1}\cdots\otimes v_r\otimes\omega^1\otimes\cdots\otimes\omega^{k-1}\otimes\omega^{k+1}\cdots\otimes\omega^s   \\
+        & \qquad  =: \omega^k(v_h)\, v_1\otimes\cdots\otimes \widehat{v}_{h} \otimes \cdots\otimes v_r\otimes\omega^1\otimes\cdots\otimes\widehat{\omega}^{k}\otimes\cdots\otimes\omega^s.
   \end{align}
 \end{notation}
 
@@ -319,8 +319,8 @@ \section{Tensors}
   \end{equation}
   By definition, the $(1,2)$-contraction is
   \begin{align}
-    \widetilde\tau(\nu^1,w_1,w_2) & = \tau(e^i,\nu^1,w_1,e_i,w_2) \\
-                                  & = e^i(v_1)\;\nu^1(v_2)\omega^1(w_1)\omega^2(e_i)\omega^3(w_2) \\
+    \widetilde\tau(\nu^1,w_1,w_2) & = \tau(e^i,\nu^1,w_1,e_i,w_2)                                                                                        \\
+                                  & = e^i(v_1)\;\nu^1(v_2)\omega^1(w_1)\omega^2(e_i)\omega^3(w_2)                                                        \\
                                   & = \LaTeXunderbrace{e^i(v_1)\omega^2(e_i)}_{=\omega^2_i e^i (v_1) =\omega^2(v_1)}\nu^1(v_2)\omega^1(w_1)\omega^3(w_2) \\
                                   & = \omega^2(v_1)\; v_2\otimes\omega^1\otimes\omega^3 (\nu^1,w_1,w_2).
   \end{align}
@@ -427,10 +427,10 @@ \section{Tensor bundles}
   \end{equation}
   This immediately exposes the transformation laws for the change of coordinates: let $(U, \psi)$ be another chart on $U$ with local coordinates $(y^i)$, then $dy^i = \psi^* de^i$ and $\frac{\partial}{\partial y^i} = (\psi^{-1})_* e_i$. If we denote $\sigma = \psi\circ\varphi^{-1}$ the transition map in $\R^n$, we get
   \begin{align}
-    \frac{\partial}{\partial x^i} & = (\varphi^{-1})_* e_i                                             \\
+    \frac{\partial}{\partial x^i} & = (\varphi^{-1})_* e_i                                                 \\
                                   & = (\varphi^{-1})_* \LaTeXunderbrace{(\sigma^{-1})_*\sigma_*}_{\id} e_i \\
                                   & = (\varphi^{-1}\circ \sigma^{-1})_* (\sigma_* e_i)                     \\
-                                  & = (\psi^{-1})_* ((D\sigma)_i^j e_j)                                  \\
+                                  & = (\psi^{-1})_* ((D\sigma)_i^j e_j)                                    \\
                                   & = (D\sigma)_i^j \frac{\partial}{\partial y^j},
   \end{align}
   which may be easier to think about in terms of the following diagram
@@ -590,9 +590,9 @@ \section{Tensor bundles}
     g := \sum_{i\in I} \rho_i \hat{g}_i,
     \quad\mbox{where}\quad
     \hat{g}_i := \begin{cases}
-      \varphi_i^* g_{\R^m} & \mbox{on } U_i, \\
-      0 & \mbox{otherwise}
-    \end{cases},
+                   \varphi_i^* g_{\R^m} & \mbox{on } U_i,  \\
+                   0                    & \mbox{otherwise}
+                 \end{cases},
   \end{align}
   concluding the proof.
 \end{proof}

6-differentiaforms.tex

@@ -286,7 +286,7 @@ \section{Differential forms on manifolds}
   \begin{equation}
     \Omega^*(M) = \bigoplus_{k=0}^n \Omega^k(M),
   \end{equation}
-  then the wedge product turns $\Omega^*(M)$ into an associative, anticommutative graded algebra.
+  then the wedge product turns $\Omega^*(M)$ into an associative, anticommutative graded algebra\footnote{Recall Remark~\ref{rem:gradedtensoralgebra}}.
 \end{remark}
 
 The following theorem gives a computational rule for pullbacks of differential forms similar to the ones we developed for covector fields and arbitrary tensor fields earlier.