- #1
caffeinemachine
Gold Member
MHB
- 816
- 15
Let $f:\mathbf R^n\times \mathbf R^k\to \mathbf R^n\times \mathbf R^k$ be a diffeomorphism such that:
1) $f(\{\mathbf p\}\times \mathbf R^k)=\{\mathbf p\}\times \mathbf R^k$, for all $\mathbf p\in \mathbf R^n$.
2) $f|\{\mathbf p\}\times\mathbf R^k:\{\mathbf p\}\times \mathbf R\to \{\mathbf p\}\times \mathbf R^k$ is a vector space isomorphism.
To show that there exists $\tau:\mathbf R^n\to GL_k(\mathbf R)$ such that $f(\mathbf p,\mathbf v)=(\mathbf p,\tau_{\mathbf p}(\mathbf v))$ for all $(\mathbf p,\mathbf v)\in \mathbf R^n\times \mathbf R^k$.
Further, $\tau$ is smooth.The exsitence of $\tau$ is immediate. What is not trivial here is the smoothness of $\tau$. Here is my attempt.
Fix the standard basis for $\mathbf R^k$.
For each $\mathbf p\in \mathbf R^n$, let $\tau_{\mathbf p}=[a_{ij}^{\mathbf p}]$ with respect to the standard basis.
Then we will be done if we can show that $\mathbf p\mapsto a_{ij}^{\mathbf p}$ is a smooth function, for each $1\leq i,j\leq k$.
We know by hypothesis that the function which sends $(\mathbf p,\mathbf v)$ to $(\sum_{j=1}^ka_{1j}^{\mathbf p}v_j,\ldots,\sum_{j=1}^ka_{kj}^{\mathbf p}v_j)$ is a smooth function.
Thus, for a fixed $i$, $(\mathbf p,\mathbf v)\mapsto \sum_{j=1}^ka_{ij}^{\mathbf p}v_j$ is smooth.
Fix $j$. Setting $v_1=\cdots=v_{j-1}=v_{j+1}=\cdots=v_k=0$, and $v_j=1$, we see that
$\mathbf p\mapsto a_{ij}^{\mathbf p}$ is smooth and hence each $a_{ij}^{\mathbf p}$ is smooth.
Is the above reasoning correct?
Also, can somebody give a coordinate free approach?
1) $f(\{\mathbf p\}\times \mathbf R^k)=\{\mathbf p\}\times \mathbf R^k$, for all $\mathbf p\in \mathbf R^n$.
2) $f|\{\mathbf p\}\times\mathbf R^k:\{\mathbf p\}\times \mathbf R\to \{\mathbf p\}\times \mathbf R^k$ is a vector space isomorphism.
To show that there exists $\tau:\mathbf R^n\to GL_k(\mathbf R)$ such that $f(\mathbf p,\mathbf v)=(\mathbf p,\tau_{\mathbf p}(\mathbf v))$ for all $(\mathbf p,\mathbf v)\in \mathbf R^n\times \mathbf R^k$.
Further, $\tau$ is smooth.The exsitence of $\tau$ is immediate. What is not trivial here is the smoothness of $\tau$. Here is my attempt.
Fix the standard basis for $\mathbf R^k$.
For each $\mathbf p\in \mathbf R^n$, let $\tau_{\mathbf p}=[a_{ij}^{\mathbf p}]$ with respect to the standard basis.
Then we will be done if we can show that $\mathbf p\mapsto a_{ij}^{\mathbf p}$ is a smooth function, for each $1\leq i,j\leq k$.
We know by hypothesis that the function which sends $(\mathbf p,\mathbf v)$ to $(\sum_{j=1}^ka_{1j}^{\mathbf p}v_j,\ldots,\sum_{j=1}^ka_{kj}^{\mathbf p}v_j)$ is a smooth function.
Thus, for a fixed $i$, $(\mathbf p,\mathbf v)\mapsto \sum_{j=1}^ka_{ij}^{\mathbf p}v_j$ is smooth.
Fix $j$. Setting $v_1=\cdots=v_{j-1}=v_{j+1}=\cdots=v_k=0$, and $v_j=1$, we see that
$\mathbf p\mapsto a_{ij}^{\mathbf p}$ is smooth and hence each $a_{ij}^{\mathbf p}$ is smooth.
Is the above reasoning correct?
Also, can somebody give a coordinate free approach?