Universal property for finite-dimensional Clifford algebras

Here’s my definition of a Clifford algebra:

Definition: Let $B(cdot,cdot)$ be a symmetric bilinear form on a vector space $V$ over $mathbb{K}$ and $Q$ its associated quadratic form. The Clifford algebra associated to the quadratic space $(V,Q)$ is a pair $mathcal{C}l(V,Q)$ where $mathcal{C}l(V,Q)$ is a $mathbb{K}$ associative algebra with identity $1$ and $varphi: V to mathcal{C}l(V,Q)$ is a Clifford map satisfying the following properties:

(a) $varphi(u)varphi(v)+varphi(v)varphi(u) = 2B(u,v)1$ for every $u,v in V$.

(b) The subspace $text{Im}varphi$ generates the algebra $mathcal{C}l(V,Q)$.

(c) For every Clifford map $phi: V to mathcal{A}$ on $(V,Q)$ there exists a homomorphism of algebras $f: mathcal{C}l(V,Q) to mathcal{A}$ such that $phi = fcirc varphi$.

Suppose $V$ is finite dimensional, with dimension $n$. Then $mathcal{C}l(V,Q)$ has dimension $2^{n}$ and is generated by $1$, $varphi(x_{i})$, $varphi(x_{i})varphi(x_{j})$, ($i<j$),…, $varphi(x_{1})cdots varphi(x_{n})$, where ${x_{1},…,x_{n}}$ is a basis for $V$. Moreover, one can prove that the Clifford map $varphi$ is injective. Thus, we may identify $V$ with the image of $varphi$, so that $V$ can be treated as a subspace of $mathcal{C}l(V,Q)$. Thus, $varphi(x_{i})$ becomes simply $x_{i}$.

In the literature, we often find the definition of a Clifford algebra with dimension $2^{n}$ as an algebra generated by (the generators) $1$, $x_{i}$, $x_{i}x_{j}$ ($i<j$),…,$x_{1}cdots x_{n}$, and these elements satisfy:
$$x_{i}x_{j} + x_{j}x_{i} = 2delta_{ij}$$
Let’s call this definition our second version of a Clifford algebra.

As I stressed before, this second case is a simplified version of the above defition, in which $V$ has dimension $n$. The generators $1$, $x_{i}$, $x_{i}x_{j}$ ($i< j$),…, $x_{1}cdots x_{n}$ are identifications of $1$, $varphi(x_{i})$, $varphi(x_{i})varphi(x_{j})$, ($i<j$),…, $varphi(x_{1})cdots varphi(x_{n})$ and $B$ is a bilinear form in which the basis of $V$ is orthonormal.

In summary, the second version follows from the first. But taking the second version as the definition of a Clifford algebra seems a little odd to me because it does not mention property (c) of the above definition, which plays an important role in the abstract theory of Clifford algebra. So my question is: does property (c) holds trivially in the case of finite dimensional vector spaces $V$, so it does not have to be demanded in the definition of such a Clifford algebra, as the second version seems to imply? If not, why is this property commonly omitted in so many references?

banach spaces – Finite-dimensional subspaces of $l_{p}$ and $c_{0}$

Let $M$ be a finite-dimensional subspace of $X=l_{p}$ or $c_{0}$. Let $epsilon>0$. Does there exist a projection from $X$, of norm $leq 1+epsilon$, onto a subspace $N$ of $X$ with $Msubseteq N$ and the Banach-Mazur distance $textrm{d}(N,l_{p}^{n})$(resp.$textrm{d}(N,l_{infty}^{n}))leq 1+epsilon$, where $n=textrm{dim}N$ ?

Thank you!

Are there examples of finite-dimensional complex non-semisimple non-commutative symmetric Frobenius algebras?

Given any finite dimensional algebra $A$, consider the linear dual $hat{A}= hom(A, k)$ as an $A$$A$-bimodule. Then $R = A oplus hat{A}$ may be equipped with an algebra structure as follows:

$$(a, x) cdot (b,y) = (ab, x cdot b + a cdot y)$$

for $a,b in A$ and $x,y in hat{A}$. The algebra $R$ has a natural symmetric Frobenius algebra structure. So every finite dimensional (possibly non-commutative) algebra embeds into a symmetric Frobenius algebra.

linear algebra – How can we make precise the notion that a finite-dimensional vector space is not canonically isomorphic to its dual via category theory?

There are quite a few questions both on this site and math.SE related to this topic as well as what we mean when we say “natural” or “canonical”. For the purposes of this question, I’m going to consider canonical and natural to be synonyms, and use wikipedia’s definition of an unnatural isomorphism:

A particular map between particular objects may be called an unnatural isomorphism (or “this isomorphism is not natural”) if the map cannot be extended to a natural transformation on the entire category.

From here: https://en.wikipedia.org/wiki/Natural_transformation

There’s an excellent question here: https://math.stackexchange.com/q/622589/816960 which raises the issue that it seems to be an artefact of the definition of a natural transformation that there is no canonical isomorphism between $V$ and $V^*$, since the dual functor is contravariant. One of the answers suggests that the only way to resolve this is by showing that every dinatural transformation from the identity functor to the dual functor must be zero.

Personally I feel this doesn’t get around the issue, because it seems to require us to redefine a canonical isomorphism between objects in a category as one which can be extended to a nontrivial dinatural transformation, and abandon the definition given above. Otherwise we’re still left with a “definition not applicable”-based proof.

Another candidate is given here: https://mathoverflow.net/a/345148/175537 in which we change the definition of the dual functor.

One way to get around this is by working instead with the core groupoid $mathbf{Vect}_{core}$, consisting of vector spaces and invertible linear transformations, and defining $*:mathbf{Vect}_{core} to mathbf{Vect}_{core}$ to be the functor taking $f:V to W$ to $(f^{-1})^{ast}: V^ast to W^ast$, the linear adjoint of its inverse. Then one can ask whether the identity is naturally isomorphic to the covariant dual functor $ast$. It is not.

But it seems like in both cases, we need a different definition of something to prove the nonexistence of a canonical isomorphism.

My questions are:

Can we prove that any isomorphism between a finite-dimensional vector space and its dual is unnatural using the definition above, without appealing to the fact that the definition of a natural transformation doesn’t allow comparisons of covariant functors with contravariant ones?

Since I suspect the answer to the above is “no”, does this mean the definition of “unnatural isomorphism” isn’t quite right? what is the right definition of “canonical isomorphism” to use in order to do this properly with category-theoretic machinery?

fa.functional analysis – Infinite-dimensional Gaussian measure vs finite-dimensional Wiener measure

I’m trying to figure out the connections between two contructions of Gaussian measure.

Let $(U, langlecdot,cdotrangle_U)$ be a seprable Hilbert space, and $mathcal{B}(U)$ be the Borel sigma-algebra.

Definition 2.1, page 10 gave the following definition of Gaussian measure:

A measure $mu$ on $(U, mathcal{B}(u), mu)$ is Gaussian if for all $vin U$, there are $m_vin mathbb{R}$, $sigma_vin mathbb{R}^+$ such that if $sigma_v>0$,

mu(u in Ucolon langle v,urangle_U in A) = frac{1}{sqrt{2pisigma_v}}int_Aexp(-frac{(s-m_v)^2}{2sigma_v^2})ds,

for all $Ainmathcal{B}(mathbb{R})$.

However, I recall from when we construct the Brownian motion. See, page 23 of https://www.springer.com/gp/book/9780387287201

Let $U$ be the space of continuous funcitons $u(t)$ on $(0,1)$, and $D$ be a cylindrical set

D = {uin U: (u(t_1), u(t_2), ldots, u(t_N))in Ein mathcal{B}(mathbb{R}^N)}

Then the measure is defined by

mu(D) = int_E prod_{i=1}^N (frac{1}{sqrt{2pi(t_i-t_{i-1})}} exp(-frac{(h_i-h_{i-1})}{2(t_i-t_{i-1})}))dh_1dh_2cdots dh_N

on $0<t_1<t_2<cdots<t_N<1$.

My questions is: Is it possible to define the infintite Gaussian measure on finite-dimensional distributions similar to we did for Brownian motion? I don’t know how to find the connection bewtween the inner product $langle v,urangle_U$ and a finite-dimensional distribution.

real analysis – Finite-dimensional unitary operators

I need help in proving the following

Let $A$ and $B$ be finite-dimensional unitary operators. Then either
$$ {left|(frac{A+B}{2} )^k right|}<1 $$
for all $k geq 1$ , or else for some vector $f ne 0$ we have
$$Af=Bf; A^2f=ABf=BAf=B^2 f; A^3f=A^2Bf=ABAf=ยทยทยท=B^3f; … $$

I know that the norm of unitary operator is $1$,and if ${left|f+g right|}={left|f right|}+{left|g right|}$ with ${left|f right|}={left|g right|}=1$ then $ f=g$ (the equality case of triangle inequality ).

linear algebra – let V be an inner product space, S and $ S_0 $ subsets of V and W a finite-dimensional subspace of V.

Let V be an inner product space, S and $ S_0 $be subsets of V and W be a finite-dimensional subspace of V.

To prove:$ S_0 subset S $ implies $ S ^ { bot} subset S_0 ^ { bot} $

Pf: Let $ x in S ^ { bot} $, for all $ y in S $, the inner product $ langle x, y rangle $ given is $ langle x, y rangle = 0 $what implies $ y in S_0 $ Then $ S_0 subset S $, $ x in S_0 ^ { bot} $Proven.

To prove: $ W = (W ^ { bot}) ^ { bot} $

Pf: Let $ x in W $ and $ y in W ^ { bot} $.
By definition the inner product is given for all y in W ^ { bot} $ rangle x, y langle = 0 $.
Use of $ x in W $, which gives $ W subset (W ^ { bot}) ^ { bot} $ How can you prove the other side?

Representation theory – convolutional algebra in connection with finite-dimensional algebra

Given a finite dimensional algebra $ A $ (We can assume that there is a connected quiver with relations). You can make your trivial extension $ T (A) $ (see for example https://math.stackexchange.com/questions/229412/trivial-extension-of-an-algebra), which is a Frobenius algebra and therefore has a coal Gebra structure, see for example the book by Kock on Frobenius algebras and topological 2D quantum field theories.
Now if we have any coalgebra C and algebra A, we can form the convolutionalgebra $ W_ {C, A} = Hom_K (C, A) $,
This leads to (at least) two new algebras from a given algebra $ A $namely $ W_ {T (A), A} $ and $ W_ {T (A), T (A)} $ in order to $ T (A) $ has a coalgebra structure as Frobenius algebra. I have no real experience with this topic. I'm sorry if my questions are stupid.

Question 1: Has one of the algebras $ W_ {T (A), A} $ or $ W_ {T (A), T (A)} $ already considered / described? Are they quiver algebra and can they be described?

Question 2: What do these two algebras look like in the specific example? $ A = kQ $ is a path algebra (possibly of the Dynkin type).

Question 3: If these questions are not trivial, can these algebras be calculated by a computer algebra system like GAP / QPA?

Representation theory – "Nice" foundations for finite-dimensional semi-simple Hopf algebras

To let $ H $ be a finite dimensional semisimple Hopf algebra about $ mathbb {C} $, Can you choose a base? $ {h_1, dots, h_n } $ from $ H $, Where $ h_1 = 1 $so if we write
Delta (h_i) = sum_ {1 leq j, k leq n} alpha_ {ijk} h_j otimes h_k

for some $ alpha_ {ijk} in mathbb {C} $, then $ alpha_ {ij1} = 0 $ for all $ i, j $, except $ i = 1 $ (as obviously $ Delta (1) = 1 x 1 $)? That is, when we write $ Delta (h_i) $ as the sum of $ n $ Tensors in which the second tensor edge is a basis vector is $ h_1 $ the only basis vector with a non-zero summand of the form $ – otimes $ 1?

If $ H = mathbb {C} G $ If it is a group algebra, then the base made up of group-like elements obviously has this property. If $ G = mathbb {Z} _p $ is cyclic in the order $ p $, then $ ( mathbb {C} mathbb {Z} _p) ^ * $ has this property (and I think that's true for any finite abelian group). Many examples I know have this property (eg the Kac-Palyutkin-Hopf algebra). However, I can not say if this is true $ ( mathbb {C} S_3) ^ * $, for example.

In general, if we leave $ G (H) $ denote the set of groups of groups, then it might be useful to look at a base that consists of associations of groups of groups of groups $ G (H) $, However, I did not manage to get this going.

linear algebra – Lax-Milgram lemma, alternative proof in finite-dimensional case

I wonder if we would weaken the assumptions of the Lax-Milgram lemma to the finite-dimensional case

Lax Milgram Lemma
To let ($ V $. $ ( cdot, cdot $. $ Vert cdot Vert $) be a (real) Hilbert space. To let $ mathcal {A}: V rightarrow V ^ * $ Be a linear, strongly positive, bounded operator, then it follows $ mathcal {A} $ is bijective, d. H. for all $ f in V ^ * $ There is a unique solution $ u in V $ to the problem $ mathcal {A} u = f $ in the $ V ^ * $,

How would the proof be simpler?
Since we are finally dimensional, we get a linear system of equations. So we have to show that $ A = mathcal {A} $ is invertible. But how can you show that? $ det A neq 0 $?

Thank you for the answers!