@leslietownes Think of this as DogAteMy's version of atonal music.

i respect any attempt to shock the bourgeoisie

ted: any fun this holiday weekend? we had our daughter up last night at 9pm to watch the city fireworks from our balcony. today, mostly chores around the house, with munchkin supervising.

Nope, no fun. Not much to celebrate! How did Munchkin sashimi go?

she liked some of it. her favorite was fatty tuna, which won't exist when she is an adult due to overfishing.

best to try it now.

we were just on quite a walk through the neighborhood. we met a cat that was being taken on a walk, and a neighbor showed us where a dove was raising two baby doves. they blinked at munchkin.

Cat, meet doves. Ooops.

this cat was no threat to doves. 15 years old, blurry eyes. the 'walk' was the neighbor going on a walk while holding her.

i do hope the coyote doesn't find the doves

Screech almost scratched my eye earlier. Fun play turns dangerous. At least no crazy with assault rifle … Sigh.

Happy 4th!

happy 4th, copper.

rather subdued 4th here. family in Paros. my in-laws are sick.

we had unexpected fireworks last night. the city scheduled them for the night of the 3rd and we could see them from our balcony.

cooked shepherds pie for myself

last night was peaceful, the night before someone set off (while driving I think) fireworks around 2am

we're having veggie burgers

i am so sorry.

i tried to entice a few folks, but they all had plans

losing my touch

lost probably

we have some white wine in the refrigerator that might salvage this

people said they had food already, which seems unlikely at 2pm on the 4th. they were just blowing me off

to be fair, the little star event was my choice of date, it could have been today.

i would like a glass, but i think it would flatten me now

i love dry white wine. it is the best

for relaxing, a nice cab for me

even though i used to belittle folks for doing so, i often add ice cubes nowadays

only to white

my dad was a red wine only guy. he loves the stuff that napa is known for, which is 14+% fruit bombs.

the lady who showed us the baby doves has an amazing cockney accent and is reliably drinking white wine with ice cubes every time we speak to her.

Cabs are too heavy for me, unless with a rich red meat (which will kill me anyhow).

I just put a bottle of dry rosé to chill.

i vacillate between cabs and pinots

cabs flatten me knowadays

I do like burgundy and bordeaux :)

ohhh, yeahh. absolutely

And sharp dry whites .

my advisor loved bordeaux.

my ideal wine these days is an austrian white, maybe 11% alcohol. more than that and i fall asleep.

i usually have a tiny glass before bed

i love the smells

instagram.com/p/CfnC0KMpT8P

promoting my cat's instagram

she slipped under 500 followers somehow. this is wrong

pure evil

my daughter's insta page had a compliment to me in the header, while understandable, it has been replaced by the rather humdrum graduation dates.

i am sure she has no idea how crestfallen i was

Screech would follow Livvy if she did social media.

i am sure animatronic upgrades for cats are in our future.

fields medals announced!

the article about the drop out poet in Quanta Magazine is worth a look @CalvinKhor

@copper.hat couldn't the graduation dates have been added instead of replacing the compliment?

my daughter provoked the cat this morning until she got a paw in the face. then she acted like it was the end of the world.

then, having learned nothing, she picked the cat up and wandered around the house with her.

it's a wonder that human children survive in the wild...

wait a minute...

i don't think livvy would survive in the wild either. although she does give a very good interpretation of the apex predator in this house. as long as it isn't raining too hard or something else that's scary

they would if we had any nearby. but we live close enough to disneyland to hear their nightly fireworks in the distance. she's used to that.

actually, our dog does not seem to react to the fireworks

nor do our cats, but they are a bit needy this morning.

@robjohn Part of growing up, I suppose.

yeah, but when they want the keys to the car, they are your friends once again.

:-) Except in her case, she wants me to be the Chauffeur as well

@Thorgott That proof I gave for $S_{\infty}$ not being residually finite also shows that $A_{\infty}$ is not residually finite, right? In fact, it generalizes to show that any infinite simple group is not residually finite? Or am I being a knucklehead?

@robjohn Fireworks are disallowed where I am (thank goodness).

they are for the general public here as well, but organized events with permits can have shows.

This does not stop people from setting off fireworks in their front yards.

or outside my house at 2am

Yesterday, the people across the street from where we were, the neighbors were setting off huge fireworks from their driveway. One of them rained ash down on us.

bits of burned cardboard, etc

the police were called, but they were either too busy or were just inundated with calls.

Helicopters flew overhead after all was over, of course too late to catch the culprits.

@robjohn Yeah, no personal fireworks is a pretty common bit of local and/or state legislation in the US (some states are more permissive, and some municipalities put greater limits on things). But there are not even municipal displays here---it is too dangerous (given the heat, drought, etc).

@XanderHenderson parts of LA were like that, but Burbank had their display, and Warner Center, which is in LA City, had a display.

Santa Monica had none

Huh... I would have figured that Santa Monica would do something over the water.

@user193319 it even generalizes to show any group containing an infinite simple group is not residually finite

@XanderHenderson They usually have a display on the pier, but not this year according to what I read.

It all depends on what the local politicians can pull

Indeed.

If I were inclined to watch a fireworks display, I feel like Santa Monica Pier would be a fun place for it.

long beach did one over the water

@XanderHenderson That seems like a place where it might get too crowded

I guess you could see it for a ways down the beach, but the coastline curves and there are mountains to the sea in several places that might block the view

though it's really an equivalent fact, I guess

@robjohn My assumption is that they would launch from the pier, but that folk would watch from the beach, no?

the point is that residual finiteness is an hereditary property

and infinite simple groups have trivial profinite completion, so aren't residually finite

that's probably the best way to phrase it

argument's the same one in any case

@XanderHenderson yes, that is what I would assume, as well

@robjohn Right. Yeah. Ok.

It has been a while since I've been to Santa Monica---I guess I don't have a good picture of the shape of the coast there in my head.

Pacific Palisades is an example of the mountains to the sea, but there are similar features near Santa Monica

Not that Pacific Palisades is far from Santa Monica

They announced the Fields Medal winners, including the second-ever female Fields medalist (Maryna Viazovska)

the only medal i will get in mathematics is if someone throws one at me

Apparently the founder of the place I'm gonna start working at tomorrow went to grad school with June Huh (one of the other medalists)

Oh, you have a summer job in Seattle? Cool!

No, I'm flying back to New York (boarding in a few minutes)

Oh.

The Seattle trip was visiting family

My brother-in-law is a fantastic cook. He fed us very well

You should learn to be a fantastic cook :)

It does sound useful

One day I should take a class on it

LOL ... yes. Not starving is a good use. But also it's a good way to treat your friends, etc.

Tensor product learning layman beginner here, if we have a tensor product from $ V \times W \stackrel{\tau}{\to} V \otimes W$ can we use the same product $\tau$ on two different spaces $ U \times Z \stackrel{\tau}{\to} U \otimes Z$

Or is it such that we actually get a tensor product by defining it on sets, IE it is "attached to those sets" and unique to them

I know how we can get the tensor product from the universal attribute, and that a bilinear form is attached to those sets, which make think the answer to the question is "no"

Be careful. Tensor product is not a map on $V\times W$. (Think dimensions.)

We defined $\tau$ to be the bilinar mapping from said sets to said sets as i draw them, fullfilling the attribute, and then we said, the tensor product is then a pair of the resulting vector space and the bilinear function Tau

So i am asking, if we can take this same $\tau$ and apply it to another cartesian product, or is this $tau$ as i understand, inherited with the sets

the attribute needs to be fulledfilled for each bilinear map from $V \times W \to Z $ we have only one linear function such that $b \circ \tau $ equal to that function

Thats our def

I know what the universal mapping property is. I'm just saying your sentence "tensor product from $V\times W$ to $V\otimes W$" is just wrong.

Oh OK. So i guess The tensor product defined on..?

You can map $(v,w)$ to $v\otimes w$, but the tensor product is a lot bigger than the image of your map.

Anyhow, the construction is analogous, no matter which vector spaces you start with. I don't know what the rest of your point was.

So $Im(\tau) \neq V \otimes W$?

Thats because of the "none pure" tensors right? because for pure tensors its obvious mapping using $\tau$, i guess you would call them Simple as well in English

Correct. As I said, compute dimensions.

Yes $ n+m \neq nm$ for $n,m \neq 2$

So it is atleast injective mapping however, right.

That's a good question. Can you prove it?

trying...

the thing is, every element with 0 will be zeroed. i am not sure if that is equivalent to say, that the kernel of bijective map is not trivial

i know how you would check injectivity for a linear mapping, but not bilinear.

so (0,w) , (v,0) for all v w are zeros. but does this still mean kernel is trivial. idk. for linear maps its just "zero" i am not sure if those also qualify as "zeros"

Probably not

but yea then the kernel will have dimension of n+m...so that probably means not nice.

my final answer is no

Wait thats rubbish

I just made my kernel the whole space

i dont know...

Dim kernel = 2?

Ted is appaled by my stupidity

No. But you don't have a linear mapping at all.

yes i know it is bilinear.

Does your course really work with $\tau$ and not just the universal mapping property?

Well, yea kind of..

it like changes...

I find it very confusing.

Teddy, i have no freaking clue what these objects even look like :(

i am just doing symbols

Thinking of bilinear maps is a good thing. Think of $V\otimes W$ as the vector space of bilinear maps on $V^*\times W^*$.

So it's more natural to work with $V^*\otimes W^*$ as the vector space of bilinear maps on $V\times W$.

Linear, Geometric, Bilinear, Topological, and Quasilinear Maps

So is it then not injective, how do you show this? is it long proof ?

What is "it"?

$\tau$

I'm telling you to forget about $\tau$. What's the point of it?

Okay. Forgot about it.

If my professor asks me this question, i shall refer him to you and throw the blame on you

I shall quote and say "Forget about it" :D

And it must not be injective for teh sake of argument since (v,0), (o,w) are in the kern

So if $\tau$ makes sense, then what is $\tau\big((v,w)+(v',w')\big)$?

Well, it has to be $\tau\big((v+v',w+w')\big) = (v+v')\otimes (w+w')$. It's definitely not $\tau(v,w)+\tau(v',w')$.

you are calculating this not correctly

you will get (i will just write the arguments) (v,w),(v',w'),(v,w'),(v',w)

maybe i am translating badly

The sum of those four is what I have.

well then i dont understand your senteance above.

My point, though, is that $\tau$ is definitely not linear.

Yes i know, it is linear in each component.

why are you repeating this, am i making a mistake that make it seem i am considering tau to be linear?

Trying to understand the kernel, image seems something you would do with linearity.

This is what i meant at the start...

i dont know how to check a bilinear function for injectivity

i can only do it for linear ones.

Well, you already showed it never is.

There's a notion of nondegeneracy of a bilinear form.

Anyhow, I think you should concentrate on the universal mapping property. I don't even know if your professor gave you a recipe for constructing the tensor product.

Hello

Today I met the son of the mathematician who introduced "semigroups"

is he a mathematician

must be pretty old? the notion goes back at least 100 years

unless there's some modern abstract semigroup theory i don't know about (which is possible)

no the guy I met is not a mathematician but he has a degree in mathematics

he's really an entrepreneur

@leslietownes yes he's like 55 I'm guessing