Differential Geometry – Show $ langle exp ^ {- 1} _ {x} (y), exp ^ {- 1} _ {x} (y) rangle_ {g_ {x}} = langle exp ^ {- 1} _ {y} (x), exp ^ {- 1} _ {y} (x) rangle_ {g_ {y}} $

To let $ M $ a riemannian manifold n-dimensional without boundary. Show that $$ langle exp ^ {- 1} _ {x} (y), exp ^ {- 1} _ {x} (y) rangle_ {g_ {x}} = langle exp ^ {- 1 } _ {y} (x), exp ^ {- 1} _ {y} (x) rangle_ {g_ {y}} qquad, forall y in mathcal {U} $$
Where $ mathcal {U} $ is a normal neighborhood of $ x in M ​​$

My approach: Let $ mathcal {U} $ a normal neighborhood of $ x in M ​​$that contain one $ y in M ​​$, If $ c (t) $ is the geodesic curve connecting both points ($ x, y $), i. $ c (0) = x $ and $ c (1) = y $, then $ c ^ { prime} (0) = exp_ {x} ^ {- 1} (y) $,

Now consider the parallel transport of $ c ^ { prime} (0) $ along $ c $ (Note that $ c $ is the geodesic curve), then we have $$ langle c ^ { prime} (0), c ^ { prime} (0) rangle_ {g_ {x}} = langle c ^ { prime} (1), c ^ { prime} (1) rangle_ {g_ {y}} $$

I do not know if my proof is correct *. However, why, if we take the parallel transport, $ P_ {c} rvert_ {0} ^ {1} $ along the geodesic line, the end point coincides with the point $ c (1) = y $? Is this point unique?

Every idea is appreciated. THANK YOU SO MUCH!

linear algebra – If $ v $ and $ w $ are orthogonal to each other with respect to the scalar product $ langle cdot, cdot rangle $, they are also linearly independent.

To let $ V $ an Euclidean $ mathbb {R} $Vector room with a generic scalar product $ langle cdot,
cdot rangle $
, To let $ v, w $ With $ v neq 0 $ and $ w neq 0 $, Show: When $ v $ and $ w $ are orthogonal to each other with respect to the dot product $ langle cdot,
cdot rangle $
They are also linearly independent. (It should be a general scalar product)

I'm not sure how to start. I would assume that they are linearly dependent, but that did not lead to a good solution.

To let $ v = lambda w, lambda in mathbb {R} $ and $ langle v, w rangle = 0 dots $

Is $ L = { langle M rangle mid L (M) subseteq HP } in coRE $?

My intuition is that $ L notin coRE $but I did not manage to prove it $ HP le L $, as before, I have seen only from discounts $ HP $ or from $ overline {HP} $ With $ f $ so that $ f (( langle M rangle, x)) = langle M_x rangle $while $ M_x $ performs a simulation of $ M $ on $ x $,

(The answer was hidden from me for a while, so I started to write a question here.) After finding the surprisingly simple answer, I decided to post it anyway (Q & A-style).)

Turing Machines – Checking $ L = { langle M, w, n rangle $: $ M $ accepts $ w $ within $ n $ steps $ } $ is decidable

Show the following problem is decidable: Given $ w in Sigma ^ {*} $, $ n in mathbb {N} $and a Turing machine $ M $does $ M $ on $ w $ Stop inside $ n $ Steps.

My thoughts:

I am new to testing such results with Turing machines, so often I'm not sure what is legal or not legal with a Turing machine. For example,
My textbook often contains "high-level" descriptions of Turing machines that perform a calculation rather than formally step by step. That's how we're likely to describe Turing machines in the classroom, but I'm often not sure I'm strict enough in my description.


My attempt:

Construct a Turing machine $ M & # 39; $ in the following way:

$ M & # 39; $ = "When entering $ langle M, w, n rangle $:

(1) Representation $ n $ on the tape of $ n $ consecutive $ 1 $& # 39; s

(2) simulate $ M $ on $ w $

(3) Each "execution" of the transition function is followed by a mark on one of the $ 1 $& # 39; s

(4) If $ M $ enters an accepting state, accepting; Otherwise, refuse if $ M $ enters a negative state or after all $ 1 $'S are marked. "


My main concern is step (1) of the algorithm. I'm really not sure if that's possible. I'm not sure how $ n $ is displayed as a string for input $ langle M, w, n rangle $, Can I choose how it is encrypted?

Thank you for your time and feedback.

Linear Algebra – Find the first $ 3 $ orthogonal polynomials of Laguerre (in terms of $ langle f, g rangle = int_0 ^ { infty} f (x) g (x) e ^ {- x} dx $)

Find the first one $ 3 $ orthogonal polynomials of Laguerre (in relation to
$ langle f, g rangle = int_0 ^ { infty} f (x) g (x) e ^ {- x} dx $)

I do not know what definition to use for the Laguerre polynomials, and whether my definition produces orthogonal polynomials.

I found this recursive definition:

$$ L_0 = 1, L_1 = 1-x \ $$

$$ L_ {k + 1} (x) = frac {(2k + 1-x) L_ {k} (x) -kL_ {k-1} (x)} {k + 1} $$

So I can easily get the third. However, I do not know if these polynomials are orthogonal because the Wikipedia article mentions this generalized Laguerre polynomials are orthogonal and I do not think they are generalized. Besides, I did not use the dot product from the exercise. I think my way is not what my teacher intended for me.

Shows an equivalence of $ | langle textbf {z}, x rangle | <R $

Consider the following event: for a fix $ x in mathbb {R} ^ n $ With $ | x | = 1 $ and $ textbf {z} sim N (0, textbf {I} _n) $,
$$
A_x ( textbf {z}) = { exists w in mathbb {R} ^ n text {so} | w – textbf {z} | le R, text {and} text {sign} ( langle w, x rangle) ne text {sign} ( langle textbf {z}, x rangle) }.
$$

I try to show that
$$
A_x ( textbf {z}) text {happens} iff | langle textbf {z}, x rangle | <R.
$$

($ implies $) This instruction is easily executed as follows.
Accept $ A_x ( textbf {z}) $ happens. Then there is $ w in mathbb {R} ^ n $ so that $ | w – textbf {z} | R $ and $ text {sign} ( langle w, x rangle) ne text {sign} ( langle textbf {z}, x rangle) $,
That's why,
$$
R ge | w – textbf {z} | ge | langle (w – textbf {z}), x rangle | ge | langle textbf {z}, x rangle |
$$

where the third inequality is used $ text {sign} ( langle w, x rangle) ne text {sign} ( langle textbf {z}, x rangle) $,

I'm not sure why the reverse is true.
Here is my try.

($ impliedby $) Accept $ R ge | langle textbf {z}, x rangle | $,
Then our goal is to find it $ w in mathbb {R} ^ n $ satisfying
(I) $ | w – textbf {z} | R $ and (ii) $ text {sign} ( langle w, x rangle) ne text {sign} ( langle textbf {z}, x rangle) $,
I tried to construct $ w $ by setting it $ w = -a textbf {z} $ for a positive constant $ a $, This satisfies condition (ii) automatically. And I was hoping that (i) can be satisfied by careful selection $ a $,
However, this only works under the condition of $ | textbf {z} | R $Not
$ | langle textbf {z}, x rangle | R $, This is because of the attitude $ w = -a textbf {z} $, we have
$$
| w – textbf {z} | = (1 + a) | textbf {z} |.
$$

This is less than or equal to $ R $,
$$
a + 1 le frac {R} { | textbf {z} |} iff a le frac {R | textbf {z} |} { | textbf {z} |}.
$$

Since $ a $ is positive, we need $ | textbf {z} | <R $,

Suggestions / answers / comments are highly appreciated.

Is $ M = frac { mathbb {Z} ^ 3} { langle (3,3,1), (2,2,2) rangle} $ free?

I try to solve a question that asks me to determine the quotient $ mathbb {Z} $-Module $ M = frac { mathbb {Z} ^ 3} { langle (3,3,1), (2,2,2) rangle} $ free. I should then find an element of $ M $ infinite order.

I am really new to modules and find it difficult to wrap them around my head. I would appreciate any help you could give me.

Random graphs – distribution of $ ( langle X_i, X_j rangle) _ {i, j = 1} ^ n $ for $ X_k sim operatorname {Unif} (S ^ d) $

For two independent random variables evenly distributed on a $ d $Spherical surface ($ X_1, X_2 sim operatorname {Unif} (S ^ d) $), It's obvious that
$$ langle X_1, X_2 rangle sim – langle X_1, X_2 rangle $$
by symmetry of $ X_1 $ and linearity of the inner product. This term can be generalized to the common distribution of the pairwise scalar combination $ n $ such variables through clever use of symmetries? It does
$$ ((1- delta_ {ij}) langle X_i, X_j rangle) _ {i, j = 1} ^ n sim (- (1- delta_ {ij}) langle X_i, X_j rangle ) _ {i, j = 1} ^ n $$
stop? Through rotation invasion, one can derive the density of each such entry fairly quickly, but that does not seem to be helpful because dependency / interaction is more involved.

ag.algebraic geometry – Determining the genus and the number of poles in a ring $ mathbb {C}[t,t^{-1},u]/ langle u ^ m -p (t) rangle $?

A pole of the finite polynomial $ p (t) = sum_ {i in mathbb {Z}} a_it ^ i $ is a point $ p_0 $ so that $ p (p_0) = infty $,

I know that when $ m = 2 $the genus $ g $ depends on the defining polynomial $ p (t) $ with grad $ d $ according to the formula
$$
g = frac {d-1} {2},
$$

equivalent
$$
2g = begin {cases} d-2 textrm {if $ d $ is even} \ d-1 textrm {if $ d $ is odd.} End {cases}
$$

After that I have the number $ n $ Where sticks are allowed depends on it $ p (t) $ according to the formula $ n = 4-r $ from where $ r $ is the number of ramified points in $ {0, infty } $: $ 0 $ is ramified exactly to the constant term $ a_0 = 0 $, and $ infty $ exactly when the degree ramifies $ d $ is odd. The combination of this information gives
$$
2g + n-1 = begin {cases} d + 1 textrm {if} a_0 neq0, \ d-1 textrm {if} a_0 = 0. end {cases}
$$

As I said, it is when $ m = 2 $, When $ m $ is arbitrary, what could I say?

How to determine the genus and the number of poles in a ring
$ mathbb {C}[t,t^{-1},u]/ langle u ^ m -p (t) rangle $?

linear algebra – Show that $ langle x, y rangle = 0 $.

Suppose that $ left (X, langle cdot, cdot rangle right) $ be a complex inner product space. To let $ x, y in X $ be so $ | alpha x + beta y | ^ 2 = | alpha x | ^ 2 + | beta y | ^ 2 $ for all couples $ alpha, beta in Bbb C $ $ ($ from where $ | cdot | $ is the norm that is caused by the inner product $ X $ $). $ Then $ langle x, y rangle = 0 $ i.e. $ x perp y. $

How do I show this? Please help me with this.

Many Thanks.