2,223 reputation
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bio website math.berkeley.edu/~sramesh
location Berkeley, CA
age 28
visits member for 4 years, 2 months
seen Apr 13 at 20:50
I am a graduate student in the Logic program at Berkeley, broadly interested in categorical logic and foundations of mathematics, as well as in applications of category theory to the semantics of programming languages.

Nov
5
comment What is a complex inner product space “really”?
I believe the $\omega$ is a reference to Elizabeth's answer (as is the $g$, for that matter).
Nov
4
comment What is a complex inner product space “really”?
Similarly, here is why I said $J$ has to rotate by 90 degrees: $Jv \cdot v$ is the real component of $(iv) \* v$, which by "sesquilinearity" is the real component of $i(v \* v)$, which by positive-definiteness is $0$. Thus, $Jv \cdot v = 0$. It follows from those properties that $J^2 = -1$ [consider $(x + Jx) \cdot J(x + Jx)$], but just taking $J$ to be an arbitrary linear operator satisfying $J^2 = -1$ will not, under the transformation outlined above, produce a complex inner product (with "sesquilinearity" and positive-definiteness).
Nov
4
comment What is a complex inner product space “really”?
Here is why I said $J$ has to be length preserving: Suppose you have a real inner product $\cdot$, linear operator $J$, and complex inner product $\*$, related by the correspondence in my post. Then $Jv \cdot Jv$ is the real component of $(iv) \* (iv)$, which by "sesquilinearity" is the real component of $v \* v$, which is $v \cdot v$. Thus, $Jv \cdot Jv = v \cdot v$.
Nov
4
comment What is a complex inner product space “really”?
The interesting thing to me is the question of why "real inner product space" should be a mathematically natural concept. Geometric intuition? Is there anything more to that than just "Well, as a contingent empirical matter, the universe we live in involves a real inner product space of significance"? That hardly seems mathematically satisfactory. What, on abstract, a priori grounds, makes positive definite symmetric bilinear forms so special? (The best I can come up with is that they are the multilinear-ization of {0, 1}-valued equality relations on a basis. But is there a better story?)
Nov
4
revised What is a complex inner product space “really”?
Stray J that should now be an R
Nov
4
revised What is a complex inner product space “really”?
added 454 characters in body
Nov
4
revised What is a complex inner product space “really”?
added 668 characters in body; deleted 5 characters in body
Nov
4
answered What is a complex inner product space “really”?
Nov
1
comment Can the Riemann hypothesis be undecidable?
In what way? From the quote in your answer, Martin Davis's reason seems to amount to "Brilliant mathematicians, including Paul Cohen, have tried to prove it for a while without success. Wouldn't it be great if it were independent from [whatever]?"
Nov
1
awarded  Mortarboard
Nov
1
awarded  Nice Answer
Nov
1
comment Can the Riemann hypothesis be undecidable?
Oh, but then, to know that there are NO zeros would require checking each of infinitely many domains. So, yes, this gives an algorithm to semi-decide for the existence of non-trivial zeros, which establishes the Riemann Hypothesis as a $\Pi_1$-statement (and thus disprovable if false), but doesn't rule out the possibility of undecidability altogether.
Nov
1
revised Can the Riemann hypothesis be undecidable?
added 221 characters in body
Nov
1
answered Can the Riemann hypothesis be undecidable?
Nov
1
comment Can the Riemann hypothesis be undecidable?
Are you sure those finite algorithms allow one to decide (as in, will definitely give either a "YES" or "NO" answer) whether or not there are non-trivial zeros? Or do they just semidecide (as in, will say "YES" if the answer is "YES", but run forever if the answer is "NO") or some such thing?
Oct
29
comment Number of distinct values taken by x^x^…^x with parentheses inserted in all possible ways
The same phenomenon occurs for any natural number, of course: b^(b^(b^b) * the product of b many bs) = b^(b^(the product of b many bs) * b^b). So every natural number b fails to be generic for parenthesization of x^x^x... with 4 + b many copies of x. [Paraphrased from "The Nesting and Roosting Habits of The Laddered Parenthesis", by R. K. Guy and J. L. Selfridge]
Oct
29
comment Number of distinct values taken by x^x^…^x with parentheses inserted in all possible ways
3^(3^(3^3) * 3 * 3 * 3) = 3^(3^(3 * 3 * 3) * 3^3). [I've written these using products in the exponent, which of course can be rewritten into iterated exponentiations in various equivalent orders]
Oct
14
comment Ways to prove the fundamental theorem of algebra
Alternatively presented: Consider the map from the Riemann sphere to the Riemann sphere induced by a polynomial. This is a continuous map from a compact space to a Hausdorff space, and thus its image is closed. It is also trivially holomorphic, and thus, if non-constant, its image is open. But the domain is inhabited and the codomain is connected, so this map must be surjective.
Sep
22
comment Mathematical computing rules-of-thumb
@brianjd: I see (I was thrown by the mention of vectors, though I should've seen your earlier clarification). In that case, it's surely simply for the same reason that x^2 can be slower than x * x for single element calculations: the general procedure to compute x^y with arbitrary (possibly even non-integral) exponent y is presumably rather slower than the procedure to simply multiply two values (this latter operation probably being particularly primitively implemented in hardware).
Sep
21
comment Mathematical computing rules-of-thumb
I suspect item 1 means "Compute x DOT x, instead of |x|^2", and that the reason this is faster is that the computation of |x| proceeds by computing sqrt(x DOT x)...