# 09-17 Notes

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==Sobolev norms on sections== | ==Sobolev norms on sections== | ||

− | Let $A$ be a connection in a vector bundle $E \to X$. For $\xi$ a section of $E$, we may define | + | Let $A$ be a connection in a vector bundle $E \to X$. For $\xi$ a section of $E$, we may define the Sobolev $L^p_{j,A}$ norm by $\norm{\xi}^p_{L^p_{j,A}} = \sum_{i=0}^j \int_X |\nabla_A^i\xi|^p \ d\mathrm{vol}$. The topology induced by this norm is independent of the connection $A$ in the following precise sense: If $A_0$ is a $C^\infty$ reference connection, and $A \in \mathcal{A}^p_k$ is any $L^p_k$ connection, then $L^p_{j,A}$ and $L^p_{j,A_0}$ norms are equivalent as long as the inequalities $(k+1)p > n$ and $0\leq j \leq k+1$ hold. |

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+ | To prove this, write $A = A_0 + a$. We proceed by induction on $j$. It will suffice to show that $\norm{\nabla_{A_0 + a}\xi}_{L^p_{j,A_0}} \leq \norm{\xi}_{L^p_{j+1,A_0}}(1+\norm{a}_{L^p_{k,A_0}}).$ The left hand side of this inequality is less than $\norm{\nabla_{A_0}\xi}_{L^p_{j,A_0}} + \norm{a\xi}_{L^p_{j,A_0}} \leq \norm{\xi}_{L^p_{j+1,A_0}} + \norm{a}_{L^p_{k,A_0}}\norm{\xi}_{L^p_{j+1,A_0}}.$ In this last inequality, we have applied the multiplication theorem to the second term, and this requires $(j-n/p) \leq (j+1 - n/p) + (k-n/p),$ which is the same as $(k+1)p>n$. | ||

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+ | We conclude that the $L^p_{k+1,A}$ norms are all equivalent for $A \in \mathcal{A}^p_k$. Note that the condition $(k+1)p>n$ is also the condition for gauge group $\mathcal{G}^p_{k+1} \subset L^p_{k+1}(X,\mathrm{End}(E))$ to consist of continuous sections and act continuously on $\mathcal{A}^p_k$. |

## Latest revision as of 12:05, 29 October 2008

## Sobolev norms on sections

Let $A$ be a connection in a vector bundle $E \to X$. For $\xi$ a section of $E$, we may define the Sobolev $L^p_{j,A}$ norm by $\norm{\xi}^p_{L^p_{j,A}} = \sum_{i=0}^j \int_X |\nabla_A^i\xi|^p \ d\mathrm{vol}$. The topology induced by this norm is independent of the connection $A$ in the following precise sense: If $A_0$ is a $C^\infty$ reference connection, and $A \in \mathcal{A}^p_k$ is any $L^p_k$ connection, then $L^p_{j,A}$ and $L^p_{j,A_0}$ norms are equivalent as long as the inequalities $(k+1)p > n$ and $0\leq j \leq k+1$ hold.

To prove this, write $A = A_0 + a$. We proceed by induction on $j$. It will suffice to show that $\norm{\nabla_{A_0 + a}\xi}_{L^p_{j,A_0}} \leq \norm{\xi}_{L^p_{j+1,A_0}}(1+\norm{a}_{L^p_{k,A_0}}).$ The left hand side of this inequality is less than $\norm{\nabla_{A_0}\xi}_{L^p_{j,A_0}} + \norm{a\xi}_{L^p_{j,A_0}} \leq \norm{\xi}_{L^p_{j+1,A_0}} + \norm{a}_{L^p_{k,A_0}}\norm{\xi}_{L^p_{j+1,A_0}}.$ In this last inequality, we have applied the multiplication theorem to the second term, and this requires $(j-n/p) \leq (j+1 - n/p) + (k-n/p),$ which is the same as $(k+1)p>n$.

We conclude that the $L^p_{k+1,A}$ norms are all equivalent for $A \in \mathcal{A}^p_k$. Note that the condition $(k+1)p>n$ is also the condition for gauge group $\mathcal{G}^p_{k+1} \subset L^p_{k+1}(X,\mathrm{End}(E))$ to consist of continuous sections and act continuously on $\mathcal{A}^p_k$.