Chemistry - Why do we need the identity operator, E

Solution 1:

It is not that we need the identity operator. It is just that things (character tables, irreducible representations, etc.) work the way they do. As to why they do so, I refer you to any decent textbook on group theory.

I guess one could rebuild the entire theory from scratch (that is, reformulate all definitions, starting from the very definition of a group, reprove all theorems, etc.) without explicitly using $\rm E$. That would be a titanic amount of work, though, and quite pointless at that. Also, many things like "order of a group is divisible by..." would start to sound less natural.

Come to think of it, why do we need the number $0$ in math? It just sits there and does nothing, after all. Well, doing nothing may be quite important: it makes a lot of other things simpler.

Solution 2:

What you are asking arises as a fundamental consequence of the definition of vector spaces and the operations defined for them and, specifically, symmetry groups. The identity operator does not do "nothing" (see below) and it does belong in the formal definition of a group, and it is included to satisfy said definition.

From Byron and Fuller's Mathematics of Classical and Quantum Physics:

...a group is a system consisting of a set of elements and rules for combining the members of this set.

This sets the stage for the definition:

A group is a system $\{G, \cdot\}$, consisting of the set $G$ and a single closed operation $\cdot$, which satisfies the following three axioms:

  1. If $a, b, c \in G$, then $a\cdot(b\cdot c) = (a\cdot b) \cdot c$ (associativity).

  2. There exists an identity element $e\in G$ such that for all $a\in G$, $a\cdot e = e\cdot a = a$.

  3. For every $a\in G$, there exists an inverse element in $G$, denoted $a^{-1}$, such that $a\cdot a^{-1} = a^{-1}\cdot a = e$.

Given this, we can go to more plain English as Wilson, Decius, and Cross do in Molecular Vibrations, where they note that

It is evident that for every symmetry operation possessed by a molecule there is some operation, either the same one or a different one, which undoes the work of the first.

This is an important point, and operationally illustrates what the identity operator does and why it is included in every group:

...if an operation $R$ changes a molecule from the position I to the position II, then there is an operation which shifts the molecule from II to I. Such an operation is called the inverse of the original operation and is written $R^{-1}$. Symbolically, $R^{-1}R = E$, since the symbol $E$ represents the identity operation, which leaves the molecule unaltered.

(For the record, WDC also write that "...$E$ is used for the trivial operation of leaving a molecule in its original position.")


Solution 3:

Well, according to this Math StackExchange post, you need the identity element because it is impossible to define an inverse element a if you don't know about the identity element e. The definition of a group includes the identity element (below).

There exists an element e in [Group] G such that, for every element a in G, the equation e • a = a • e = a holds. Such an element is unique (see below), and thus one speaks of the identity element. (Wikipedia)


Solution 4:

Symmetry operator $E$ is the lowest symmetry a molecule may possess; and indeed all objects possess this operator. Yet, among molecules revealing $E$ as the only the operation are the ones exhibiting stereogenic centred chirality, like $\ce{CFClBrI}$. Hence $E$ is of value, not redundant, and needed.


Solution 5:

This is a special case of a phenomenon that occurs throughout mathematics. For example, you could ask why we have zero for similar reasons. The answer is frequently: it's often easier to reason about a collection of things if the thing that represents 'nothing' is considered part of the collection.

For example, if we didn't have zero there'd be two kind of sentences about how much money we have: "I have \$X" or "I don't have any money". If you allow X to be zero then you only need one kind of sentence "I have \$X".

In the case of groups, many theorems are easier to state if you include the identity as part of your collection. For example "the order of a subgroup divides the order of a group" is much easier than "one plus the order of a subgroup divides one plus the order of a group". And "in a group you can always multiply two elements" becomes "you can multiply elements most of the time but there are certain pairs you can't multiply because when you multiply them they cancel out doing nothing but we don't consider doing nothing to be an element of the group". Closer to your example, the statement of the orthogonality relations for character tables would be more complex.

The fact that so many things simplify when you include the identity is good reason for it being there.

Tags:

Group Theory