Invariant characterization of isometric embeddings

Question

$\newcommand{\Cof}{\operatorname{Cof}}$ $\newcommand{\Det}{\operatorname{Det}}$ $\newcommand{\Lam}{\operatorname{\Lambda}}$

Motivation (and the "classic" case):

I am trying to find a coordinate-free criterion to determine when a linear map is an isometric embedding of an inner product space in a space of higher dimension.

Such a criterion exists when the dimensions are equal. Let $V,W$ be oriented inner product spaces, of the same dimension $d$. Let $\,A: V \to W$ be a linear map. Then one can define invariantly the cofactor and the determinant of $A$ as follows: $$(1) \Det A := \star_W^d \circ \bigwedge^d A \circ \star_V^0 ,$$ $$(2) \Cof A = (-1)^{d-1} \star_W^{d-1} \circ \bigwedge^{d-1} A \circ \star_V^1,$$

Note that $\Det A \in \mathbb{R}$,$\, \Cof A:V \to W$, (I keep the degree explicit in the notation of the Hodge-dual, i.e $\star_V^p:\Lam_p(V) \to \Lam_{d-p}(V)$).

The well-known matrix identity $$\Cof X \cdot X^T = \det X \cdot \operatorname{Id}$$ can be invariantly written as $$(*) \, \Cof A \circ A^T= \Det A \cdot \operatorname{Id}_W$$

Immediate observation: $(*)$ implies that $A$ is an isometry if and only if $$(3) \Det A \neq 0 \text{ and} \, \Det A \cdot A= \Cof A$$

My question is about the validity of a possible approach of generalizing criterion $(3)$ to the case where $\dim V < \dim W$.

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The case $\dim V < \dim W$:

Stage I: generalizing identity $(*)$.

Assume $\dim V=d < n=\dim W$. Definitions $(1),(2)$ for the cofactor and the determinant now gives objects of different types:

(a) $\Cof A:V \to \bigwedge^{n-d+1}W$, so $\Cof A \circ A^T:W \to \bigwedge^{n-d+1}W$.

(b) $\Det A \in \bigwedge^{n-d}W$.

The natural generalization of identity $(*)$ should therefore be:

$$(**) \Cof A \circ A^T(w)= \Det A \wedge w \, ,\,\forall w \in W.$$ (This is an equality of elements in $\bigwedge^{n-d+1}W $).

Stage II: generalizing condition $(3)$.

I think $A$ will be an isometric embedding if and only if $$ (3)' \Cof A \, \text{is injective and} \, \Det A \wedge A(v) = \Cof A(v) \, \, \forall v \in V $$ (This is an equality of elements in $\bigwedge^{n-d+1}W $).

Indeed, putting $w=Av$ in $(**)$, one gets that for any linear map $A$, $$ \Cof A \circ A^T(Av)= \Det A \wedge Av, $$ so $A$ satisfies condition $(3)'$ if and only if $$ \Cof A \, \text{is injective and} \, \Cof A \circ (A^TA) = \Cof A$$ which holds iff $A^TA=\operatorname{Id}_V$ (which is equivalent to $A$ being an isometry).

The remaining questions (which are needed for completion of this endeavour) are:

(1) Does identity $(**)$ hold? (It probably holds "up to sign", i.e both $d,n$ should enter the new definition of the cofactor in a manner which should be determined )

(2) In both cases, part of the conditions were "non-triviality" , in $(3)$ it was $\det A \neq 0$ which is equivalent to the injectivity of $A$. In $(3)'$ it was injectivity of $\Cof A:V \to \bigwedge^{n-d+1}W$. Is it true that injectivity of $A$ implies injectivity of $\Cof A$? (and what about the converse?)

Answer 1

Here is a proof of $\left( \ast\ast\right) $. As you correctly admitted, part of the problem is to define the signs that enter into $\operatorname*{Cof}A$. Let me actually start from scratch and introduce all notations that will be needed:

For any $m\in\mathbb{N}$, we shall write $\left[ m\right] $ for the set $\left\{ 1,2,\ldots,m\right\} $.

Let $d$ and $n$ be two nonnegative integers. Let $A\in\mathbf{k}^{n\times d}$ be an $n\times d$-matrix over a commutative ring $\mathbf{k}$. Regard $A$ as a $\mathbf{k}$-linear map $\mathbf{k}^{d}\rightarrow\mathbf{k}^{n}$.

Let $\left( f_{1},f_{2},\ldots,f_{d}\right) $ be the standard basis of the $\mathbf{k}$-module $\mathbf{k}^{d}$. (Thus, $f_{i}$ is the vector whose $i$-th entry is $1$ and whose all other entries are $0$.)

Let $\left( e_{1},e_{2},\ldots,e_{n}\right) $ be the standard basis of the $\mathbf{k}$-module $\mathbf{k}^{n}$. For every subset $I=\left\{ i_{1} <i_{2}<\cdots<i_{k}\right\} $ of $\left[ n\right] $, we let $e_{I}$ be the vector $e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k}}\in\wedge ^{k}\left( \mathbf{k}^{n}\right) $.

If $U\subseteq\left[ n\right] $ and $V\subseteq\left[ d\right] $, then we let $A_{U,V}$ denote the submatrix of $A$ obtained by removing all rows except for those in $U$ and all columns except for those in $V$. (In more rigorous terms: We let $A_{U,V}$ be the matrix $\left( a_{u_{i},v_{j}}\right) _{1\leq i\leq\left\vert U\right\vert ,\ 1\leq j\leq\left\vert V\right\vert }$, where $A$ is written in the form $\left( a_{i,j}\right) _{1\leq i\leq n,\ 1\leq j\leq d}$, and where the sets $U$ and $V$ are written in the forms $U=\left\{ u_{1}<u_{2}<\cdots<u_{\left\vert U\right\vert }\right\} $ and $V=\left\{ v_{1}<v_{2}<\cdots<v_{\left\vert V\right\vert }\right\} $.)

If $S$ is a finite set of integers, then $\sum S$ shall mean the sum of the elements of $S$.

Regard the $d\times n$-matrix $A^{T}\in\mathbf{k}^{d\times n}$ as a $\mathbf{k}$-linear map $\mathbf{k}^{n}\rightarrow\mathbf{k}^{d}$.

Define an element $\operatorname*{Det}A\in\wedge^{n-d}\left( \mathbf{k} ^{n}\right) $ by $\operatorname*{Det}A=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d}}\left( -1\right) ^{\sum I-\left\vert I\right\vert }\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}$.

Define a $\mathbf{k}$-linear map $\operatorname*{Cof}A:\mathbf{k} ^{d}\rightarrow\wedge^{n-d+1}\left( \mathbf{k}^{n}\right) $ by setting

$\left( \operatorname*{Cof}A\right) \left( f_{i}\right) =\sum _{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1}}\left( -1\right) ^{i+\sum J}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) e_{J}$.

Theorem 1. For every $w\in\mathbf{k}^{n}$, we have $\left( \left( \operatorname*{Cof}A\right) \circ A^{T}\right) \left( w\right) =\left( \operatorname*{Det}A\right) \wedge w$ in $\wedge\left( \mathbf{k} ^{n}\right) $.

Proof of Theorem 1. Let $w\in\mathbf{k}^{n}$. We must prove the equality $\left( \left( \operatorname*{Cof}A\right) \circ A^{T}\right) \left( w\right) =\left( \operatorname*{Det}A\right) \wedge w$. This equality is $\mathbf{k}$-linear in $w$; thus, we can WLOG assume that $w$ is an element of the basis $\left( e_{1},e_{2},\ldots,e_{n}\right) $ of $\mathbf{k}^{n}$. In other words, we can WLOG assume that $w=e_{k}$ for some $k\in\left[ n\right] $. Assume this, and consider this $k$.

Write the matrix $A$ in the form $A=\left( a_{i,j}\right) _{1\leq i\leq n,\ 1\leq j\leq d}$. Thus, $A^{T}=\left( a_{j,i}\right) _{1\leq i\leq d,\ 1\leq j\leq n}$.

From $w=e_{k}$, we obtain $A^{T}\left( w\right) =A^{T}\left( e_{k}\right) =\sum_{i\in\left[ d\right] }a_{k,i}f_{i}$ (since $A^{T}=\left( a_{j,i}\right) _{1\leq i\leq d,\ 1\leq j\leq n}$). Now,

$\left( \left( \operatorname*{Cof}A\right) \circ A^{T}\right) \left( w\right) $

$=\left( \operatorname*{Cof}A\right) \underbrace{\left( A^{T}\left( w\right) \right) }_{=\sum_{i\in\left[ d\right] }a_{k,i}f_{i}}=\left( \operatorname*{Cof}A\right) \left( \sum_{i\in\left[ d\right] }a_{k,i} f_{i}\right) $

$=\sum_{i\in\left[ d\right] }a_{k,i}\underbrace{\left( \operatorname*{Cof} A\right) \left( f_{i}\right) }_{\substack{=\sum_{\substack{J\subseteq \left[ n\right] ;\\\left\vert J\right\vert =n-d+1}}\left( -1\right) ^{i+\sum J}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) e_{J}\\\text{(by the definition of }\operatorname*{Cof}A\text{)}}}$

$=\sum_{i\in\left[ d\right] }a_{k,i}\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1}}\left( -1\right) ^{i+\sum J}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) e_{J}$

$=\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1}}\left( \sum_{i\in\left[ d\right] }\left( -1\right) ^{i+\sum J}a_{k,i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) \right) e_{J}$

$=\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1}}\left( -1\right) ^{\sum J}\left( \sum_{i\in\left[ d\right] }\left( -1\right) ^{i}a_{k,i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) \right) e_{J}$

(1) $=\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1;\\k\in J}}\left( -1\right) ^{\sum J}\left( \sum _{i\in\left[ d\right] }\left( -1\right) ^{i}a_{k,i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) \right) e_{J}$

$+\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1;\\k\notin J}}\left( -1\right) ^{\sum J}\left( \sum_{i\in\left[ d\right] }\left( -1\right) ^{i}a_{k,i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) \right) e_{J}$

(here we have split the sum into two subsums, according to whether $k\in J$ or $k\notin J$).

If $J$ is a finite set of integers, and if $k$ is an integer, then $\#\left( J<k\right) $ shall denote the number of all elements of $J$ that are smaller than $k$. Similarly, $\#\left( J>k\right) $ shall denote the number of all elements of $J$ that are greater than $k$.

But every $\left( n-d+1\right) $-element subset $J$ of $\left[ n\right] $ satisfying $k\in J$ satisfies

(2) $\sum_{i\in\left[ d\right] }\left( -1\right) ^{i+\#\left( \left[ n\right] \setminus J<k\right) +1}a_{k,i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) =\det\left( A_{\left( \left[ n\right] \setminus J\right) \cup\left\{ k\right\} ,\left[ d\right] }\right) $

(by Laplace expansion, applied to the matrix $A_{\left( \left[ n\right] \setminus J\right) \cup\left\{ k\right\} ,\left[ d\right] }$, with respect to the $\#\left( \left[ n\right] \setminus J<k\right) +1$-th row of this matrix (which used to be the $k$-th row of $A$)). Hence, every $\left( n-d+1\right) $-element subset $J$ of $\left[ n\right] $ satisfying $k\in J$ satisfies

(3) $\sum_{i\in\left[ d\right] }\left( -1\right) ^{i}a_{k,i} \det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus \left\{ i\right\} }\right) =\left( -1\right) ^{\#\left( \left[ n\right] \setminus J<k\right) +1}\det\left( A_{\left( \left[ n\right] \setminus J\right) \cup\left\{ k\right\} ,\left[ d\right] }\right) $

(this follows by multiplying both sides of (2) with $\left( -1\right) ^{\#\left( \left[ n\right] \setminus J<k\right) +1}$).

Furthermore, every $\left( n-d+1\right) $-element subset $J$ of $\left[ n\right] $ satisfying $k\notin J$ satisfies

(4) $\sum_{i\in\left[ d\right] }\left( -1\right) ^{i}a_{k,i} \det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus \left\{ i\right\} }\right) =0$

(by Laplace expansion, applied to the matrix $A_{\left[ n\right] \setminus J,\left[ d\right] }$ with an extra copy of the $k$-th row of $A$ added to its top; the result is $0$ because the resulting matrix has this $k$-th row appear twice inside it). Now, (1) becomes

$\left( \left( \operatorname*{Cof}A\right) \circ A^{T}\right) \left( w\right) $

$=\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1;\\k\in J}}\left( -1\right) ^{\sum J}\underbrace{\left( \sum _{i\in\left[ d\right] }\left( -1\right) ^{i}a_{k,i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) \right) }_{\substack{=\left( -1\right) ^{\#\left( \left[ n\right] \setminus J<k\right) +1}\det\left( A_{\left( \left[ n\right] \setminus J\right) \cup\left\{ k\right\} ,\left[ d\right] }\right) \\\text{(by (3))}}}e_{J}$

$+\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1;\\k\notin J}}\left( -1\right) ^{\sum J}\underbrace{\left( \sum _{i\in\left[ d\right] }\left( -1\right) ^{i}a_{k,i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) \right) }_{\substack{=0\\\text{(by (4))}}}e_{J}$

$=\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1;\\k\in J}}\underbrace{\left( -1\right) ^{\sum J}\left( -1\right) ^{\#\left( \left[ n\right] \setminus J<k\right) +1}}_{=\left( -1\right) ^{\sum J+\#\left( \left[ n\right] \setminus J<k\right) +1}}\det\left( A_{\left( \left[ n\right] \setminus J\right) \cup\left\{ k\right\} ,\left[ d\right] }\right) e_{J}$

$=\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1;\\k\in J}}\left( -1\right) ^{\sum J+\#\left( \left[ n\right] \setminus J<k\right) +1}\det\left( A_{\left( \left[ n\right] \setminus J\right) \cup\left\{ k\right\} ,\left[ d\right] }\right) e_{J}$

$=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d;\\k\notin I}}\left( -1\right) ^{\sum\left( I\cup\left\{ k\right\} \right) +\#\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) <k\right) +1}\underbrace{\det\left( A_{\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) \right) \cup\left\{ k\right\} ,\left[ d\right] }\right) }_{\substack{=\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) \\\text{(since }\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) \right) \cup\left\{ k\right\} =\left[ n\right] \setminus I\text{)} }}\underbrace{e_{I\cup\left\{ k\right\} }}_{\substack{=\left( -1\right) ^{\#\left( I>k\right) }e_{I}\wedge e_{k}\\\text{(because the sign of the permutation}\\\text{that takes }e_{I\cup\left\{ k\right\} }\text{ to } e_{I}\wedge e_{k}\text{ is }\left( -1\right) ^{\#\left( I>k\right) }\text{)}}}$

(here, we have substituted $I\cup\left\{ k\right\} $ for $J$ in the sum, because every $\left( n-d+1\right) $-element subset $J$ of $\left[ n\right] $ satisfying $k\in J$ must have the form $I\cup\left\{ k\right\} $ for some $\left( n-d\right) $-element subset $I$ of $\left[ n\right] $ satisfying $k\notin I$)

$=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d;\\k\notin I}}\left( -1\right) ^{\sum\left( I\cup\left\{ k\right\} \right) +\#\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) <k\right) +1}\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) \left( -1\right) ^{\#\left( I>k\right) }e_{I}\wedge e_{k}$

$=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d;\\k\notin I}}\underbrace{\left( -1\right) ^{\sum\left( I\cup\left\{ k\right\} \right) +\#\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) <k\right) +1+\#\left( I>k\right) } }_{\substack{=\left( -1\right) ^{\sum I-\left\vert I\right\vert }\\\text{(since }\sum\left( I\cup\left\{ k\right\} \right) +\#\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) <k\right) +1+\#\left( I>k\right) \\\equiv\underbrace{\sum\left( I\cup\left\{ k\right\} \right) }_{=\sum I+k}+\#\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) <k\right) +1-\underbrace{\#\left( I>k\right) }_{\substack{=\left\vert I\right\vert -\#\left( I<k\right) \\\text{(since }k\notin I\text{)}}}\\=\sum I+k+\#\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) <k\right) +1-\left\vert I\right\vert +\#\left( I<k\right) \\=\sum I+k+1-\left\vert I\right\vert +\underbrace{\#\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) <k\right) +\#\left( I<k\right) }_{=k-1}\\=\sum I+k+1-\left\vert I\right\vert +k-1=\sum I+2k-\left\vert I\right\vert \\\equiv\sum I-\left\vert I\right\vert \operatorname{mod}2\text{)}}}\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}\wedge e_{k}$

$=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d;\\k\notin I}}\left( -1\right) ^{\sum I-\left\vert I\right\vert } \det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}\wedge e_{k}$.

Comparing this with

$\underbrace{\operatorname*{Det}A}_{=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d}}\left( -1\right) ^{\sum I-\left\vert I\right\vert }\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}}\wedge\underbrace{w}_{=e_{k}}$

$=\left( \sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d}}\left( -1\right) ^{\sum I-\left\vert I\right\vert } \det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}\right) \wedge e_{k}$

$=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d}}\left( -1\right) ^{\sum I-\left\vert I\right\vert }\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}\wedge e_{k}$

$=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d;\\k\notin I}}\left( -1\right) ^{\sum I-\left\vert I\right\vert } \det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}\wedge e_{k}$

(here, we have removed all addends which satisfy $k\in I$, because all such addends contain the vanishing factor $e_{I}\wedge e_{k}=0$), we should obtain $\left( \left( \operatorname*{Cof}A\right) \circ A^{T}\right) \left( w\right) =\left( \operatorname*{Det}A\right) \wedge w$. This proves Theorem 1.