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Wedge products and exterior powers are discussed in W. Greub's book Multilinear algebra as follows.

Definition: Let $E$ be an arbitrary vector space and $p \ge 2$. Then a vector space $\bigwedge^{p}E$ together with a skew-symmetric $p$-linear map $\bigwedge^{p}: E\times \cdots \times E \to \bigwedge^{p}E$ is called a $p$-th exterior power of $E$ if the following conditions are satisfied:

(1) The vectors $\bigwedge^{p}(x_{1},\dotsc,x_{p})\mathrel{:=} x_{1}\wedge \dotsb \wedge x_{p}$ generate $\bigwedge^{p}E$.

(2) If $\psi$ is any skew-symmetric $p$ linear mapping of $\overbrace{E\times \dotsb \times E}^{\text{$p$ times}}$ into an arbitrary vector space $F$, then there exists a linear map $f\colon \bigwedge^{p}E \to F$ such that $\psi = f\circ \bigwedge^{p}$.

Now, we set: \begin{equation} \bigwedge E \mathrel{:=} \bigoplus_{n=0}^{\infty} \bigwedge^{p}E, \tag{1}\label{1} \end{equation} where $\bigwedge^{0}E \mathrel{:=} \mathbb{C}$ and $\bigwedge^{1}E \mathrel{:=} E$.

Identifying each $\bigwedge^{p}E$ with its image under the canonical injection $i_{p}\colon\bigwedge^{p}E \to \bigwedge E$, we can write $\bigwedge E = \sum_{p=0}^{\infty}\bigwedge^{p}E$. In other words, elements of $\bigwedge E$ can be thought as sequences $(v_{0},v_{1},\dotsc)$ where $v_{p} \in \bigwedge^{p}E$ for each $p\in \mathbb{N}$. Furthermore, there is a uniquely determined multiplication on $\bigwedge E$ such that the following rules hold: \begin{gather*} (x_{1}\wedge \cdots \wedge x_{p})(x_{p+1}\wedge \dotsb \wedge x_{p+q}) = x_{1}\wedge \cdots \wedge x_{p+q} \\ 1(x_{1}\wedge \cdots \wedge x_{p}) = (x_{1}\wedge \dotsb \wedge x_{p})1 = x_{1}\wedge \cdots \wedge x_{p}. \end{gather*} This turns $\bigwedge E$ into an algebra, which is called exterior (or Grassmann) algebra.

Note that Greub's construction considers arbitrary vector spaces, so that, in particular, we can take $E$ to be infinite dimensional.

Grassmann algebras are used by physicists to study fermionic systems. While searching for some material on Grassmann algebras of infinite dimensional vector spaces, I found lecture notes Fermionic functional integrals and the renormalization group by Feldman, Knörrer and Trubowitz, which has an appendix (page 75) on this topic. Their construction seems interesting, but I'm having trouble trying to relate it with Greub's construction.

The first part of their notes discusses Grassmann algebras of finite dimensional vector spaces. Then, the cited appendix starts with the statement that in order to further generalize it to infinite dimensional vector spaces we need to add a topology on these spaces. This seem not to be necessary in the general case, since Greub's construction does not consider topological vector spaces. However, I think they might have physical motivations in which the addition of a topology might be important. Their construction is as follows.

Let $I$ be a countable set. The Grassmann algebra will be generated by vector from the vector space: $$E\mathrel{:=} \ell^{1}(I)\mathrel{:=}\{\alpha\colon I \to \mathbb{C}\mathrel: \sum_{i\in I}\lvert a_{i}\rvert < +\infty\}.$$ $E$ is a Banach space with norm $\|\alpha\| \mathrel{:=}\sum_{i\in I}\lvert a_{i}\rvert$. Let $\mathcal{J}$ be the set of all finite subsets of $I$, including the empty set. Take $$\mathcal{U}(I) = \ell^{1}(\mathcal{J}) \mathrel{:=}\{\alpha\colon \mathcal{J} \to \mathbb{C}\mathrel: \sum_{I\in \mathcal{J}}\lvert a_{I}\rvert<+\infty\}$$ where $a_{I} \mathrel{:=} a_{i_{1}}\dotsb a_{i_{p}}$, $I=\{i_{1},...,i_{p}\}$. Then $\mathcal{U}(I)$ is a Banach space with norm $\|\alpha\| = \sum_{I\in \mathcal{J}}\lvert a_{I}\rvert$ and, when equipped with the product: $$(\alpha \beta)_{I} \mathrel{:=}\sum_{J\subset I} \operatorname{sign}(J, I\setminus J)\alpha_{J}\beta_{I\setminus J},$$ it becomes an algebra which is called the Grassmann algebra.

With all this being said, let me get to the questions.

Feldman, Knörrer and Trubowitz's construction might not be the most general construction there is (I don't know actually, but I think it's not as I justified before). However, I'd expect their construction to be at least a particular case of Greub's general construction. However, I don't seem to be able to relate these two since the definition of $\mathcal{U}(I)$ strongly depends on its topology. So is the second construction a particular case of the first one? If not, why not? Does it have to do with the hypothesis of $E$ to be a topological vector space? Does the topology on $E$ change the definitions of objects used on Greub's construction?

NOTE: When I ask "does the topology on $E$ changes the definitions of objects on Greub's constructions?", I mean the following. If $E$ is a vector space, $\bigoplus_{n=0}^{\infty}E$ is the space of all sequences $x=(x_{0},x_{1},\dotsc)$, $x_{i} \in E$, with all but finitely many nonzero entries. If $E= \mathcal{H}$ is a Hibert space, on the other hand, $\bigoplus_{n=0}^{\infty}\mathcal{H}$ is the space of sequences with $\|x\|^{2}:=\sum_{n=0}^{\infty}\| x_{i}\|^{2}_{\mathcal{H}}<+\infty$. Thus, although $\mathcal{H}$ is itself a vector space, the norm on $\mathcal{H}$ allows us to define the direct sum in alternative way. In other words, the topology on $\mathcal{H}$ makes the difference when we define direct sums. Maybe the use of Banach spaces by Feldman, Knörrer and Trubowitz implies some modifications like this, say, to define the direct sum (\ref{1}) in an alternative way, so these two constructions might be isomorphic or something like this.

ADDED: Does anyone know this particular construction from Feldman, Trubowitz and Knörrer? Any references on this approach would be really appreciated!

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    $\begingroup$ You might be misinterpreting the notation. The $\bigoplus_{n=0}^\infty E$ that you mention as the algebraic direct sum is not the same object as the "infinite sum of Hilbert spaces". The point is that the usual direct sum construction, when applied to a sequence of complete normed spaces, usually does not yield a complete normed space. So if you want your algebraic constructions to start with complete things and produce complete things, they need to be modified $\endgroup$
    – Yemon Choi
    Jun 23 '20 at 16:24
  • $\begingroup$ @YemonChoi I think I get it. So, in summary, in the second construction I need to consider the space of sequences such that $||x||=\sum_{n=0}^{\infty}|x_{n}| <+\infty$, right? But if (\ref{1}) is modified to account to topological spaces $E$ in this way, does Greub's construction remain the same and equivalent to the second one? $\endgroup$
    – MathMath
    Jun 23 '20 at 16:38
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    $\begingroup$ TeX notes: $\wedge$ \wedge is the binary operation; you want $\bigwedge$ \bigwedge for the unary prefix operation, like $\bigwedge E$. Also, \mbox doesn't size well in superscripts, whereas \text does: compare $\overbrace{E \times \dotsb \times E}^{\mbox{$p$ times}}$ (with \mbox) to $\overbrace{E \times \dotsb \times E}^{\text{$p$ times}}$ (with \text). I have edited accordingly. $\endgroup$
    – LSpice
    Jun 23 '20 at 22:37
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    $\begingroup$ @LSpice thanks! These are very good tips!! $\endgroup$
    – MathMath
    Jun 24 '20 at 0:15
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Focusing on exterior powers here is a distraction. The main problem already appears when considering the tensor algebra $T(E)=\oplus_{n\ge 0}E^{\otimes n}$. Once the issue is understood for the tensor algebra, figuring out what to do for the exterior or symmetric algebras (e.g., Fermion or Boson Fock spaces) is trivial, because we are in characteristic zero. In positive characteristic, this becomes subtle as can be seen for example in the recent work "Koszul modules and Green’s conjecture" by Aprodu et al. where a positive characteristic Hermite Reciprocity map is constructed.

Given a vector space $E$, the first step is to consider tensor products like $E^{\otimes n}$. This can be done algebraically as in the mentioned book by Greub. However, when $E$ is an infinite dimensional TVS (topological vector space) the resulting algebraic tensor product $E\otimes\cdots\otimes E$ is a rather unsuitable object for the purposes of analysis. One typically needs to enlarge this space using a completion procedure (topology is essential for that), and one then obtains a topological tensor product $E\widehat{\otimes}\cdots\widehat{\otimes}E$. The caveat is: even when working with Banach spaces, there are lots of ways of doing that. This was Alexander Grothendieck's Ph.D. thesis work. He considered a dozen or so inequivalent reasonable definitions for these completions/versions of the tensor product which depend on the topological structure. In other words, in the course of his explorations Grothendieck found Hell. Luckily, he kept exploring and he eventually also found Paradise: the class of nuclear spaces for which all these different constructions become the same and therefore acquire a cananical feel to them.

Likewise, for the sum $\oplus_{n\ge 0}$ one typically starts with the algebraic direct sum (only finite sums allowed, i.e., we look at almost finite sequences where after a while all the terms are zero) and one then enlarges the space by taking a completion.

The construction by Feldman, Knörrer and Trubowitz is an explicit way (just a choice that works for their purposes) of doing a succession of algebraic constructions followed by topological completions, as explained above.

Now one might think that the algebraic construction as in Greub's book is more general/powerful/etc. than the topological procedure. This is a misconception. For infinite dimensional spaces that are not too big, one could in fact argue the opposite is true. Take for example the simplest infinite dimensional space: $E=\oplus_{n\ge 0}\mathbb{R}$ which can be viewed as the space of almost finite sequences of real numbers, or the space of polynomials in one variable with real coefficients. Then $T(E)$ constructed algebraically à la Greub is a particular case of the topological completion construction. Indeed, equip $E$ with the locally convex topology defined by the set of all seminorms on $E$. This is also called the finest locally convex topology. With this topology, the space is nuclear in the sense of Grothendieck's general definition (but not nuclear in the sense of the more restrictive definition used by the Russian school around Gel'fand et al., namely, the notion of countably Hilbert nuclear spaces). So that is a good sign: pretty much any reasonable completion will give you the same $E\widehat{\otimes}\cdots\widehat{\otimes}E$ which will also coincide with the algebraic tensor product (without hats). Finally for the sum one has several possible choices, but one of them will give the algebraic construction. Let us say that a seminorm on the algebraic direct sum $T(E)$ is admissible if and only if it restricts to a continuous seminorm on each summand. Take the locally convex topology on $T(E)$ defined by the set of all admissible seminorms. Take the completion. This will give nothing new. Note that all seminorms are admissible for the case $E=\oplus_{n\ge 0}\mathbb{R}$ but I wanted to introduce a more general construction which can be applied for example to $E=\mathscr{S}(\mathbb{R})$, the Schwartz space of rapidly decaying smooth functions. Then the $T(E)$ will be isomorphic as a TVS to $\mathscr{D}(\mathbb{R})$, the space of compactly supported smooth functions.

Moral(s) of the story:

For infinite dimensional spaces ordinary bases (Hamel bases) are no good. You need Schauder bases which allow infinite linear combinations. You will need to base your construction on topology. Even when topology seems to be absent, and one uses purely algebraic direct sums and tensor products, topology is still there hiding behind the scenes as in the $E=\oplus_{n\ge 0}\mathbb{R}$ example.

Recommended reading:

The excellent vignette "Schwartz kernel theorems, tensor products, nuclearity" by Paul Garrett.


July 2020 Edit:

Let me give more details on the relation between the above general methodology and the particular FKT construction. First some notation: I will write $\mathbb{N}=\{0,1,2,\ldots\}$, and I will denote the set functions from the set $X$ to the set $Y$ by $\mathscr{F}(X,Y)$. We start from the $\ell^1$ space $E$ defined as the set of functions $f\in\mathscr{F}(\mathbb{N},\mathbb{C})$ such that $$ ||f||_E:=\sum_{i\in\mathbb{N}}|f(i)| $$ is finite.

The first step is to understand the algebraic tensor product $E\otimes E$. The general construction proceeds via the free vector space with basis indexed by symbols $f\otimes g$ with $f,g\in E$ and quotienting by relations $(f_1+f_2)\otimes g-f_1\otimes g-f_2\otimes g$ etc. Another equally uninspiring construction is to take an uncountable Hamel basis $(e_i)_{i\in I}$, for $E$, produced by the Axiom of Choice, and realize $E\otimes E$ as the subset of $\mathscr{F}(I\times I,\mathbb{C})$ made of functions of finite support (equal to zero except for finitely many elements of $I\times I$). The proper definition is as a solution to a universal problem: $E\otimes E$ together with a bilinear map $\otimes:E\times E\rightarrow E\otimes E$ must be such that for every vector space $V$ and bilinear map $B:E\times E\rightarrow V$, there should exist a unique linear map $\varphi:E\otimes E\rightarrow V$ such that $B=\varphi\circ\otimes$. One can construct such a space more concretely as follows.

Let $E_2$ be the subset of $\mathscr{F}(\mathbb{N}^2,\mathbb{C})$ made of functions $h:(i,j)\mapsto h(i,j)$ which are finite sums of functions of the form $f\otimes g$ with $f,g\in E$. Here $f\otimes g$ is the function $\mathbb{N}^2\rightarrow\mathbb{C}$ defined by $$ (f\otimes g)(i,j)=f(i)g(j) $$ for all $i,j\in \mathbb{N}$. Note that the definition I just gave also provides us with a bilinear map $\otimes:E\times E\rightarrow E_2$.

Proposition 1: The algebraic tensor product of $E$ with itself can be identified with $E_2$.

The proof relies on the following lemmas.

Lemma 1: For $p,q\ge 1$, suppose $e_1,\ldots,e_p$ are linearly independent elements in $E$ and suppose $f_1,\ldots,f_q$ are also linearly independent elements in $E$. Then the $pq$ elements $e_a\otimes f_b$ are linearly independent in $E_2$.

Proof: Suppose $\sum_{a,b}\lambda_{a,b}e_a\otimes f_b=0$ in $E_2$. Then $\forall i,j\in\mathbb{N}$, $$ \sum_{a,b}\lambda_{a,b}e_a(i) f_b(j)=0\ . $$ If one fixes $j$, then one has an equality about functions of $i$ holding identically. The linear independence of the $e$'s implies that for all $a$, $$ \sum_{b}\lambda_{a,b}f_b(j)=0\ . $$ Since this holds for all $j$, and since the $f$'s are linearly independent, we get $\lambda_{a,b}=0$ for all $b$. But $a$ was arbitrary too, so $\forall a,b$, $\lambda_{a,b}=0$ and we are done.

Lemma 2: Let $B$ be a bilinear map from $E\times E$ into some vector space $V$. Suppose $g_k,h_k$, $1\le k\le n$ are elements of $E$ satisfying $$ \sum_{k}g_k\otimes h_k=0 $$ in $E_2$, i.e., as functions on $\mathbb{N}^2$. Then $$ \sum_k B(g_k,h_k)=0 $$ in $V$.

Proof: This is trivial if all the $g$'s are zero or if all the $h$'s are zero. So pick a basis $e_1,\ldots,e_p$ of the linear span of the $g$'s and pick a basis $f_1,\ldots,f_q$ of the linear span of the $h$'s (no Axiom of Choice needed). We then have decompositions of the form $$ g_k=\sum_a \alpha_{k,a}e_a $$ and $$ h_k=\sum_b \beta_{k,b} f_b $$ for suitable scalars $\alpha$, $\beta$. By hypothesis $$ \sum_{k,a,b}\alpha_{k,a}\beta_{k,b}\ e_a\otimes f_b=0 $$ and so $\sum_k \alpha_{k,a}\beta_{k,b}=0$ for all $a,b$, by Lemma 1. Hence $$ \sum_k B(g_k,h_k)=\sum_{a,b}\left(\sum_k \alpha_{k,a}\beta_{k,b}\right) B(e_a,f_b)=0\ . $$

Now the proof of Proposition 1 is easy. The construction of the linear map $\varphi$ proceeds as follows. For $v=\sum_{k}g_k\otimes h_k$ in $E_2$, we let $\varphi(v)=\sum_k B(g_k,h_k)$. This is a consistent definition because if $v$ admits another representation $v=\sum_{\ell}r_{\ell}\otimes s_{\ell}$, then $$ \sum_k g_k\otimes h_k\ +\ \sum_{\ell}(-r_{\ell})\otimes s_{\ell}=0 $$ and Lemma 2 implies $$ \sum_k B(g_k,h_k)=\sum_{\ell} B(r_{\ell},s_{\ell})\ . $$ The other verifications that $E_2$ with $\otimes$ solve the universal problem for the algebraic tensor product pose no problem.

The second step is to construct a topological completion $\widehat{E}_2$ for $E_2$. I will use the projective tensor product construction $E\ \widehat{\otimes}_{\pi}E$. For $h\in E_2$, I will use the $l^1$ norm $$ ||h||_2=\sum_{(i,j)\in\mathbb{N}^2}|h(i,j)|\ . $$ I will also use the seminorm $$ ||h||_{\pi}=\inf\ \sum_k ||g_k||_E\times||h_k||_E $$ where the infimum is over all finite decompositions $h=\sum_k g_k\otimes h_k$. The projective tensor product is the completion with respect to $||\cdot||_{\pi}$. The $||\cdot||_1$ is an example of cross norm, i.e., it satisfies $||f\otimes g||_2=||f||_E\times||g||_E$. Moreover, one has the following easy result.

Proposition 2: For all $h\in E_2$, we have $||h||_2=||h||_{\pi}$.

For the proof use the cross norm property and triangle inequality for $\le$, and for the reverse inequality, approximate $h$ by the truncation where $h(i,j)$ is replaced by zero unless $i,j\le N$.

Now it is clear that the abstract topological tensor product $\widehat{E}_2$ is nothing but the familiar $\ell^1$ space of functions on $\mathbb{N}^2$. Likewise (but with heavier notations) one can construct $\widehat{E}_n=E\ \widehat{\otimes}_{\pi}\cdots\widehat{\otimes}_{\pi}E$, $n$ times, with the corresponding $\ell^1$ norm $$ ||h||_n=\sum_{(i_1,\ldots,i_n)\in\mathbb{N}^n}|h(i_1,\ldots,i_n)|\ . $$

The topological exterior power $\widehat{E}_{n,{\rm Fermi}}$ can be identified with the closed subspace of antisymmetric functions inside $\widehat{E}_n$, namely functions $h:\mathbb{N}^n\rightarrow\mathbb{C}$ which satisfy $$ h(i_{\sigma(1)},\ldots,i_{\sigma(n)})=\varepsilon(\sigma)\ h(i_1,\ldots,i_n) $$ for all $(i_1,\ldots,i_n)\in\mathbb{N}^n$ and all permutations $\sigma$. We will equip the space with restriction of the norm $||\cdot||_n$.

Now consider the algebraic direct sum $W=\oplus_{n\ge 0}\widehat{E}_{n,{\rm Fermi}}$. Given (for the moment unspecified) positive weights $w_n$, let us define the norm $$ ||H||_{\rm Big}=\sum_{n\ge 0}w_n||h_n||_n $$ where $H$ is an element of $W$ seen as an almost finite sequence $(h_0,h_1,\ldots)$ of functions in $\widehat{E}_{0,{\rm Fermi}},\widehat{E}_{1,{\rm Fermi}},\ldots$ Clearly the completion $\widehat{W}$ is obtained by removing the almost finite restriction but still requiring convergence of the sum defining $||\cdot||_{\rm Big}$. Finally, to make contact with FKT, to $H=(h_0,h_1,\ldots)\in\widehat{W}$ we associate the set function $\alpha:\mathcal{J}\rightarrow\mathbb{C}$ where $\mathcal{J}$ is the set of finite subsets of $\mathbb{N}$ (including the empty set), as follows. For $I=\{i_1,\ldots,i_n\}\in\mathcal{J}$ with $i_1<\cdots<i_n$ we let by definition $$ \alpha(I)=h_n(i_1,\ldots,i_n)\ . $$ If we pick the weights $w_n=\frac{1}{n!}$, then this correspondence is a bijective isometry with the giant $\ell^1$ space of FKT.

Remark: One can do the same long construction with $\ell^2$ norms instead of $\ell^1$ norms, and this will produce the Fermionic Fock space of the Hilbert space $\ell^2(\mathbb{N})$, as in the mathematical literature on second quantization, e.g., in the book by Reed and Simon. Note that the corresponding topological tensor products of Hilbert spaces were introduced by Murray and von Neumann in "On rings of operators", Ann. of Math. 1936, and further developed by Cook in "The mathematics of second quantization", PNAS 1951, for the needs of Quantum Field Theory.

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    $\begingroup$ And just to close the loop back to the question: To get from $T(E)$ to $\bigwedge E$ one uses the structure of a topological algebra and quotients out a closed ideal: $\bigwedge E$ is the quotient of $T(E)$ by the closure of the ideal generated by $v\otimes v$ for all $v\in E$. $\endgroup$ Jun 26 '20 at 17:11
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    $\begingroup$ Instead of quotienting, I would prefer using the symmetrizer or antisymmetrizer projections. Again we're in good shape because of the zero characteristic. $\endgroup$ Jun 26 '20 at 17:15
  • $\begingroup$ @AbdelmalekAbdesselam this was very useful! I think I'm starting to clear my thoughts now. If I may, I'd like to make some questions/observations. I'd like to focus on the explicit construction of Feldman, Trubowitz and Knörrer. If I understood it correctly, I should first think about the definitions of tensor products $\ell^{1}(I)\otimes\cdots\otimes\ell^{1}(I)$ and then figure out the exterior powers. As you pointed out, there is no recipe for defining these tensor products and since these are not nuclear (at least I think) spaces, I should basically chose one. $\endgroup$
    – MathMath
    Jun 26 '20 at 17:51
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    $\begingroup$ I see. You want to see explicitly the FKT construction as a particular case of the general method I outlined. That's a good question, but the dirty secret here is: it is probably not worth your time. You would be better off taking FKT's definition as is and use it. Working out the completion approach is rather long. For example you will have to learn en.wikipedia.org/wiki/Topological_tensor_product At least FKT work with Banach spaces so the easier notion of completion for metric spaces will suffice and you will not need... $\endgroup$ Jun 26 '20 at 18:30
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    $\begingroup$ ...en.wikipedia.org/wiki/… I don't have time now, but if I do later I will try to explain how to relate FKT to my more general answer. $\endgroup$ Jun 26 '20 at 18:32

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