How to only display the mantissa of a number (without the exponent part)

You can extract the components (mantissa or exponent) using TeX parameter text matching:

\documentclass{article}

\makeatletter
\def\@mantissa#1e#2\relax{#1}
\def\@exponent#1e#2\relax{#2}
\newcommand{\mantissa}[1]{\expandafter\@mantissa#1\relax}
\newcommand{\exponent}[1]{\expandafter\@exponent#1\relax}
\makeatother

\begin{document}
\setlength{\parindent}{0pt}% Just for this example

\newcommand{\Value}{2.36e6}

Value: \Value

Mantissa: \mantissa{\Value}

Exponent: \exponent{\Value}

\end{document}

The "non-user level" macros \@mantissa and \@exponent expects a parameter text of the form <m>e<e>\relax where e and \relax are required in the input stream. The user level macros \mantissa and \exponent both pass \relax, but that means you can only pass numbers/values that uses scientific notation, and therefore necessarily have the notation <m>e<e>.

You can extract both the mantissa and the exponent:

\documentclass{article}

\makeatletter
\newcommand{\mantissa}[1]{%
  \hafid@mantissa#1ee\@nil
}
\def\hafid@mantissa#1e#2e#3\@nil{%
  #1%
}
\newcommand{\exponent}[1]{%
  \hafid@exponent#1ee\@nil
}
\def\hafid@exponent#1e#2e#3\@nil{%
  #2%
}

\begin{document}

Mantissa: $\mantissa{1}$; Exponent: $\exponent{1}$\par
Mantissa: $\mantissa{1.2}$; Exponent: $\exponent{1.2}$\par
Mantissa: $\mantissa{1.2e3}$; Exponent: $\exponent{1.2e3}$\par
Mantissa: $\mantissa{1.2e-3}$; Exponent: $\exponent{1.2e-3}$\par

\end{document}

These macros return nothing if the mantissa or the exponent are not present. An easy modification can return 0 if the exponent is not explicitly given: just replace the definition of \hafid@exponent with

\def\hafid@exponent#1e#2e#3\@nil{%
  \ifx\relax#2\relax
    0%
  \else
    #2%
  \fi
}

A version that also allows symbolic input (a control sequence that eventually expands to a number; the exponent can also be input as E.

\documentclass{article}
\usepackage{xparse}

\ExplSyntaxOn
\DeclareExpandableDocumentCommand{\mantissa}{m}
 {
  \hafid_mantissa:f { \tl_lower_case:n { #1 } }
 }

\cs_new:Nn \hafid_mantissa:n
 {
  \__hafid_mantissa:www #1 ee \q_stop
 }
\cs_generate_variant:Nn \hafid_mantissa:n { f }

\cs_new:Npn \__hafid_mantissa:www #1 e #2 e #3 \q_stop
 {
  #1
 }

\DeclareExpandableDocumentCommand{\exponent}{m}
 {
  \hafid_exponent:f { \tl_lower_case:n { #1 } }
 }

\cs_new:Nn \hafid_exponent:n
 {
  \__hafid_exponent:www #1 ee \q_stop
 }
\cs_generate_variant:Nn \hafid_exponent:n { f }

\cs_new:Npn \__hafid_exponent:www #1 e #2 e #3 \q_stop
 {
  \tl_if_empty:nTF { #2 } { 0 } { #2 }
 }
\ExplSyntaxOff

\begin{document}

Mantissa: $\mantissa{1}$; Exponent: $\exponent{1}$\par
Mantissa: $\mantissa{1.2}$; Exponent: $\exponent{1.2}$\par
Mantissa: $\mantissa{1.2e3}$; Exponent: $\exponent{1.2e3}$\par
Mantissa: $\mantissa{1.2e-3}$; Exponent: $\exponent{1.2e-3}$\par
Mantissa: $\mantissa{1.2E-3}$; Exponent: $\exponent{1.2E-3}$\par

\newcommand{\mynumber}{1.26e10}

Mantissa: $\mantissa{\mynumber}$; Exponent: \exponent{\mynumber}

\end{document}

Note: due to a misfeature in the current release of expl3 (release 2016/01/19 r6377), the code doesn't work as intended under XeLaTeX or LuaLaTeX. The simplest workaround is not allowing E as a divider:

\DeclareExpandableDocumentCommand{\mantissa}{m}
 {
  \hafid_mantissa:f { #1 }
 }
\DeclareExpandableDocumentCommand{\mantissa}{m}
 {
  \hafid_exponent:f { #1 }
 }

Just for the sake of variety, here's a LuaLaTeX-based solution.

The following code sets up two Lua functions, named get_mant and get_expo, which extract the mantissa and the exponent, respectively, from a term such as -1.23e+45. The code also provides two TeX "wrapper" macros, named \GetMant and \GetExpo, which invoke the Lua functions. Thus, \GetMant{2.36e6} returns 2.36, and \GetExpo{2.34e-56} returns -56.

It is assumed that the inputs of \GetMant and \GetExpo are either valid numbers expressed in "scientific" notation or macros that evaluate to valid scientific-notation numbers. The code doesn't perform any input sanity checking, though: It assumes that you won't be trying to extract mantissas and exponents from expressions such as abcdefghi or +55+e-66-d. The main formatting requirement is that the input string contains one (and only one) instance of the letter e; E is OK too.

Two comments:

• The code is set up to suppress any + symbols that may be present at the start of the mantissa and exponent components. If you do want to preserve the + symbols, just remove the tonumber function calls in the arguments of tex.sprint.

• The code can handle the case of an "empty mantissa", i.e., an expression such as e22: In such an event, the mantissa is set to 1. Similarly, the code can also handle the case of an "empty exponent", i.e., an expression such as 1.6e: the exponent is set to 0. Even the extreme (and frankly somewhat absurd) empty mantissa/empty exponent case, viz., e or E, is handled correctly: the mantissa is set to 1 and the exponent is set to 0. :-)

% !TEX TS-program = lualatex
\documentclass{article}
\usepackage{booktabs}

%% Lua-side code
\usepackage{luacode}
\begin{luacode}
function get_mant ( s )
  mant = string.gsub ( s, "^(.-)[eE](.-)$", "%1" ) 
  if mant == "" then mant = "1" end
  return tex.sprint ( tonumber(mant) )
end
function get_expo ( s )
  expo = string.gsub ( s, "^(.-)[eE](.-)$", "%2" )
  if expo == "" then expo = "0" end
  return tex.sprint ( tonumber(expo) )
end
\end{luacode}

%% TeX-side code
\newcommand\GetMant[1]{\directlua{get_mant(\luastring{#1})}}
\newcommand\GetExpo[1]{\directlua{get_expo(\luastring{#1})}}

\begin{document}

\newcommand{\NumA}{-3.45e-678}
\newcommand{\NumB}{+1.23E+45}
\newcommand{\NumC}{e}

$\begin{array}{@{}lll@{}}
\mbox{Input} & \mbox{Mantissa} & \mbox{Exponent}\\
\midrule
\texttt{\NumA} & \GetMant{\NumA} & \GetExpo{\NumA}\\
\texttt{\NumB} & \GetMant{\NumB} & \GetExpo{\NumB}\\
\texttt{e22}   & \GetMant{e22}   & \GetExpo{e22}  \\
\texttt{1.23E} & \GetMant{1.23E} & \GetExpo{1.23E}\\
\texttt{\NumC} & \GetMant{\NumC} & \GetExpo{\NumC}\\
\end{array}$

\end{document}