bio | website | plusepsilon.de |
---|---|---|
location | Uni Hamburg | |
age | 30 | |
visits | member for | 4 years, 10 months |
seen | Jan 31 at 13:20 | |
stats | profile views | 8,563 |
Apr
11 |
comment |
What is the intuition behind the definition of cuspidal representations?
"some and therefore every" |
Apr
10 |
comment |
What is the intuition behind the definition of cuspidal representations?
Yes the unipotent radiacal of any Borel subgroup (defined over $F$). Note that your are allowed to conjugate by elements of $GL_2(F)$. There are may expositions on how to move between classical and adelic language, see e.g. Bump's book. |
Apr
10 |
comment |
leading-order behaviour of riemann zeta function?
$\zeta$ gets arbitrary small in $1/2 \leq \Re s <1$. I am not sure about $0< \Re s <1/2$. |
Apr
10 |
comment |
leading-order behaviour of riemann zeta function?
I am not getting it? Is your "I'm looking for something...." not stronger then LH, which is certainly not known? |
Apr
10 |
reviewed | Approve multiplication of two ergodic and stationary processes |
Apr
10 |
answered | Looking for paper: Weil's original 1952 “Sur les formules explicites de la théorie des nombres premiers” |
Apr
10 |
reviewed | Approve Does Cauchy continuity imply uniform continuity? [No.] |
Apr
10 |
comment |
Reference for Kronecker-Weyl theorem in full generality
Sorry, I didn't see the part: " a proof under the assumption that the θj are linearly independent over the rational numbers will not suffice for me." I deleted my answer. |
Apr
8 |
comment |
Inequality for a gamma function
The logarithmic derivative of the Selberg Zeta function grows like $CT^2$ as $\Im z = T \rightarrow \infty$, which can be seen from the Weyl law. More important for its growth is the Barnes-G-function. $\Gamma$ contributes at most $T \log(T)$ in the non-compact setting. |
Apr
8 |
answered | What is the intuition behind the definition of cuspidal representations? |
Apr
7 |
comment |
What is the logarithmic derivative of an (intertwining) operator?
Note that my computation apply only to the highest type. For the computations at the real places and the unramified complex cases, you can have a look at my PhD thesis. There is a good reason for working with highest types/smallest weights = irreducible $K$-reps as soon as you have pinned down the local conditions. |
Apr
6 |
comment |
Homeomorphisms that admit a decomposition
Okay, my mistake;) I see now that it seems to more complicated... |
Apr
5 |
comment |
Homeomorphisms that admit a decomposition
There are only two strictly monotone functions on $[0,1]$ up to conjugation by homeomorphisms. One doesn meet your criteria. |
Apr
5 |
comment |
Homeomorphisms that admit a decomposition
Do you know the answer for $[0,1]$? Modulo conjugating by homeomoprhism of $[0,1]$, there seems to be only one map. |
Apr
5 |
comment |
Space with 720° / not 2$\pi$ rotational symmetry?
Perhaps the Möbiusband? |
Mar
29 |
revised |
What is the logarithmic derivative of an (intertwining) operator?
added 8 characters in body |
Mar
29 |
answered | Current Status on Langlands Program |
Mar
29 |
comment |
Current Status on Langlands Program
I would also distinguish between functoriality (maps) and correspondence (objects)... you seem to be interested in correspondence between automorphic reps and Galois reps. |
Mar
29 |
revised |
What is the logarithmic derivative of an (intertwining) operator?
added 217 characters in body |
Mar
29 |
comment |
Current Status on Langlands Program
I also think that stabilization is not an issue for GL(n). |