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Version 1 |
| Let $f: S^n \to S^n$ be a continuous map. Here $n$ is a non-negative integer. Applying the $n^{th}$ reduced homology functor $\widetilde{H}_n(\_)$, we obtain a homomorphism $f_*: \widetilde{H}_n(S^n) \to \widetilde{H}_n(S^n)$. Since $\widetilde{H}_n(S^n) \approx \mathbb{Z}$, it follows that $f_*$ is a homomorphism $\mathbb{Z} \to \mathbb{Z}$. Such a map must be multiplication by an integer $d$. We define the \emph{degree} of the map $f$, to be this $d$. |
Let $f: S^n \to S^n$ be a continuous map. Here $n$ is a non-negative integer. Applying the $n^{th}$ reduced homology functor $\widetilde{H}_n(\_)$, we obtain a homomorphism $f_*: \widetilde{H}_n(S^n) \to \widetilde{H}_n(S^n)$. Since $\widetilde{H}_n(S^n) \approx \mathbb{Z}$, it follows that $f_*$ is a homomorphism $\mathbb{Z} \to \mathbb{Z}$. Such a map must be multiplication by an integer $d$. We define the \emph{degree} of the map $f$, to be this $d$. |
| \subsection{Basic Properties} |
\subsection{Basic Properties} |
| \begin{enumerate} |
\begin{enumerate} |
| \item If $f,g : S^n \to S^n$ are continuous, then deg($f \circ g$) = deg($f$)$\cdot$deg($g$). |
\item If $f,g : S^n \to S^n$ are continuous, then deg($f \circ g$) = deg($f$)$\cdot$deg($g$). |
| \item If $f,g : S^n \to S^n$ are homotopic, then deg($f$) = deg($g$). |
\item If $f,g : S^n \to S^n$ are homotopic, then deg($f$) = deg($g$). |
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\item The degree of the identity map is $+1$.
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\item The degree of the identity map is +1.
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\item The degree of the constant map is $0$.
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\item The degree of the constant map is .
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\item The degree of a reflection through an $(n+1)$-dimensional hyperplane through the origin is $-1$.
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\item The degree of a reflection through an $(n+1)$-dimensional hyperplane through the origin is -1.
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| \item The degree of the antipodal map, sending $x$ to $-x$ has degree $(-1)^{n+1}$. |
\item The degree of the antipodal map, sending $x$ to $-x$ has degree $(-1)^{n+1}$. |
| \end{enumerate} |
\end{enumerate} |
| \subsection{Examples} |
\subsection{Examples} |
| If we identify $S^1 \subset \mathbb{C}$, then the map $f : S^1 \to S^1$ defined by $f(z) = z^k$ has degree $k$. It is also possible, for any positive integer $n$, and any integer $k$, to construct a map $f: S^n \to S^n$ of degree $k$. |
If we identify $S^1 \subset \mathbb{C}$, then the map $f : S^1 \to S^1$ defined by $f(z) = z^k$ has degree $k$. It is also possible, for any positive integer $n$, and any integer |
| Using degree, one can prove several theorems, including the so-called 'Hairy-Ball Theorem', which states that there exists a continous non-zero vector field on $S^n$ if and only if $n$ is odd. |
opriate multiples of the $i$-th column of |
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$M_{ij}^{*}$ to its other columns, we can obtain a matrix with 0 at positions |
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$(j,k)$ ($k \neq i$) and 1 at position $(j,i)$. Now we apply the permutation |
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\[(1 2) \circ (2 3) \circ\dots \circ ((i-1) i)\] |
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to columns and |
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\[(1 2) \circ (2 3) \circ\dots \circ ((j-1) j)\] |
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to rows of this matrix. This results in the following arrangement: |
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\begin{itemize} |
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\item |
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The first row contains the vector $e_1$; |
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\item |
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rows 2 to $j$ contain the block of rows 1 to $j -1$ of $M_{ij}^*$; |
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\item |
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columns 2 up to $i$ below row no. 1 contain the block of columns 1 to $i-1$ of |
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$M_{ij}^{*}$. |
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\item |
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column 1 is the vector $e_1$. |
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\end{itemize} |
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Furthermore, the sign of the determinant of $M_{ij}^*$ has changed $i +j -2$ |
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times. So we have |
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\[\det{M_{ij}^*} =(-1)^{i+j}\det{M_{ji}}.\;\quad\] |
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Note also that in the computation of the determinant of $M_{ij}^*$ only those |
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permutations $\pi \in S_n$ are effective where $\pi(i)=j$. |
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Now we start out with |
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\begin{eqnarray*} |
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\det{M} &=\sum_{\pi \in S_n} \mathrm{sgn}\pi.\left(\prod_{j=1}^n |
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m_{j\pi(j)}\right) \\ |
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& =\sum_{k=1}^n m_{ik}\left(\sum_{\pi \in S_n, \pi(i)=k} |
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\mathrm{sgn}\pi\left(\prod_{1 \le j <i} m_{j\pi(j)}\right).1.\left(\prod_{i < j \le |
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n} m_{j\pi(j)}\right)\right). |
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\end{eqnarray*} |
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The interior sum associated with $m_{ik}$ is identical to the determinant of a |
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matrix obtained from $M$ by replacing its $i$-th column with the vector $e_k$ |
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(i.e., the matrix $M_{ij}^*$). So we have |
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\[\det{M} =\sum_{k=1}^n m_{ik}\det{M_{ik}^*} =\sum_{k=1}^n |
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\left(m_{ik}(-1)^{i+k}\det{M_{ki}}\right)\] |
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according to the previous lemma. |
