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| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 9 months |
| seen | Apr 18 at 16:58 | |
| stats | profile views | 30 |
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Feb 7 |
answered | Stupid question about conductors/Dirichlet Characters |
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Feb 5 |
comment |
Stupid question about conductors/Dirichlet Characters I am sorry but i do not understand what you mean. I cannot set $N$ (or $M$ as you call it), it is a fixed reference that i start with. |
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Feb 5 |
accepted | How many absolute values are there? |
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Feb 5 |
asked | Stupid question about conductors/Dirichlet Characters |
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Jan 28 |
awarded | Teacher |
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Dec 30 |
comment |
How many absolute values are there? Interesting. Thanks. |
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Dec 20 |
accepted | Reference requested: 'decomposition' of Haar measure on the adeles. |
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Dec 19 |
answered | Reference requested: 'decomposition' of Haar measure on the adeles. |
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Dec 18 |
comment |
Computing the modularity function of upper triangular matrices So, let us set $B_N := \{ x \in B_p: |det(x)| \geq p^{-N}\}$, and let $\nu$ be the restriction of the measure on $GL_2$ then we see that $B_p = \cup_{N \in \mathbb{N}} B_N$ and consequently $\nu(B_N) = \int_{GL_2(Q_p)} 1_{B_N} dx/|det(x)| \leq const \int_{GL_2(Q_p)} 1_{B_N} dx = 0$ by the last comment. $|\cdot|$ denotes the p-adic norm. |
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Dec 18 |
comment |
Computing the modularity function of upper triangular matrices No! It is not like: G an LCH group, H a subgroup then the Haar measure on H is simply the one on G restricted to H. The problem becomes visible in the above example: Every set of the form $* \times * \times \{0\} \times *$ is a Null set in $Q_p^4$ (with respect to the unique product measure $\mu$ of the usual Haar measure, remark that the cases $\infty \cdot 0$ and $0 \cdot \infty$ are defined as $0$!) So as you say: the measure on $GL_2$ is roughly $(1/det) \cdot \mu$, so $B_p$ is a Null set with respect to that restriction (see next comment). |
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Dec 17 |
asked | Product measure: why are those the two 'extreme' cases? |
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Nov 30 |
comment |
How many absolute values are there? I was very impressed by your proof, because it seems really strange to me that there is more than just the ''usual'' absolute value on $\mathbb{R}$... everything that one learned in analysis should then be doubted, because this awkward absolute value might be the ''better'' one to do analysis... |
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Nov 29 |
comment |
How many absolute values are there? I am having trouble verifying the triangle inequality for the simple case "$\alpha$ transcendental": Choose $q,w \in \mathbb{Q}$ arbitrary with $q \neq -w$ and $\beta = \alpha^0 \frac{q\alpha + 1}{1}$ and $\beta' = \alpha^0 \frac{w\alpha - 1}{1}$ then $\beta + \beta' = \alpha^1 \frac{q - w}{1}$ so their norm ist $|\beta + \beta'| = |q+w|_p$ which can be an arbitrary value but $|\beta| + |\beta'| = 1+1 = 2$. Could you check? |
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Nov 28 |
asked | How many absolute values are there? |
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Sep 24 |
asked | Computing the modularity function of upper triangular matrices |
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Sep 12 |
comment |
Open set = *disjoint* union of open balls? Ok, thanks a lot!! |
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Sep 12 |
accepted | Open set = *disjoint* union of open balls? |
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Sep 12 |
comment |
Open set = *disjoint* union of open balls? Beatiful, thank you very much!! There is only one thing that still bothers me: If one explicitely wants that balls do not have radius infinity, then one can only argue that all open sets $U \subset X$ with $U \neq X$ have disjoint coverings of open balls. In the case of $Q_p$ or more general, if the metric is discrete, open balls are also closed, so one can first remove a single open ball and then apply the above to the rest... What happens in the general case? |
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Sep 12 |
comment |
Open set = *disjoint* union of open balls? @Ahriman: Every open set in $\mathbb{R}$ is a disjoint union of open balls? Could you explain this to me? The argument seems to be the same: $\mathbb{R}$ cannot be a disjoint union of open balls since it is connected... ? |
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Sep 11 |
comment |
Open set = *disjoint* union of open balls? Side remark: what about the case $(B_n(x))_{n \in \mathbb{N}}$? This is an ascending sequence of open balls but their union is the whole space... but then the open set is also the whole space, so actually one needs to know that every ascending chain of open balls is either an open ball or the whole space :D |
